Eng Textbook

Introduction to Computers and Programmingusing

C++ and MATLAB

Alex F. BielajewThe University of Michigan

Department of Nuclear Engineering and Radiological Sciences2927 Cooley Building (North Campus)

2355 Bonisteel BoulevardAnn Arbor, Michigan 48109-2104

U. S. A.Tel: 734 764 6364Fax: 734 763 4540

email: [email protected]

c© 2000—2011 Alex F Bielajew

Revision: June 29, 2011

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Preface

This “book” arises out of a course I teach, a four-credit (52 hour), freshman-level courseIntroduction to Computers and Programming being taught in the College of Engineering atthe University of Michigan. The book is in reasonably rough shape at this stage. It wasassembled from my lecture notes several years ago and is under constant revision. I maynever finish it! A wise person once said, “Old age happens when you dwell more on the pastthan on the future.” By this definition, I have found eternal youth, insofar as this book isconcerned.

My educational objectives are quite simple. This is not really a course in computing. Itis a course in thinking, technical thinking, logical thinking, about formulating a problem,a mathematical problem, a physics problem, a game, and then, executing a solution andmaking it “work”. I consider the computer and the ability to program it as a kind oflaboratory—a laboratory to investigate practically, the theories and ideas of other technicalcourses. It is possible, for example, to teach a lot of Calculus to students without evermentioning the word, and there are several examples throughout this book.

This course is not about syntax. Hence, the book introduces the minimum amount syntaxto get through a problem. Indeed, I even keep some syntax hidden, to encourage studentsto discover their own algorithms. So, if you are thinking of using this book as a technicalreference in C++ or Matlab, I anticipate that you will be disappointed.

The greatest value in this book, if there is any to be found, is in the exercises, problems andprojects at the back of almost every chapter. The ideal way to learn a computer languageis to learn a little syntax and then try it out on a computer. The ideal way to think isnot to read about it, but to actually do it! The book reflects this. The material in thechapters, separated from the exercises, is worse than useless, for reading the material andnot doing the problems is just a waste of time. Do the problems! Moreover, don’t ask mefor the solutions! There is much more pedagogical value in a well-posed question than awell-articulated answer.

OK, I’ll step off my soap box now!

Several professors and many students have contributed to the ideas put into this book.Professor James Holloway and I have had endless discussion on the general problem ofteaching algorithmic thinking to freshman. We still have not come to any conclusions except

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that it is highly challenging and rewarding. As for how it actually gets done, well, to be sureit is a work-in-progress. I spent a very enjoyable term co-teaching this course (the first timeI taught it) with Professor Ken Powell. Thanks, Ken, for nursing me through that first year!Both James’s and Ken’s diverse computational backgrounds are reflected to some degree inthis book.

I owe a debt of gratitude to the dozens of Graduate Student Instructors who have taughtwith me on this course. One of them, Dan Osborne, deserves special recognition. He is one ofthe most gifted and committed teachers I have ever encountered. Dan and I spent hour afterhour discussing the challenges of teaching computing and thinking skills to undergraduates.

To the 5000 or so undergraduates, mostly freshmen, who have taken my course: Every timeI teach this course I learn something new—about computing, about teaching, about the joyof making an early “deflection” in a student’s career.

Finally, I am one of those people who tends to see the forest, instead of the trees. Conse-quently, spelling and sentence structure usually yield to an enthusiasm to present the materialin an interesting way. It’s not my only fault, but maybe my most visible one. To this end,please inform me if you spot any errors. I am also deeply indebted to my wife-to-be, LindaPark, for her unimaginable attention to detail, her proofreading and suggestions. Linda,every time you undertake a proofreading project on my behalf, my mind boggles, truly.

AFB, June 29, 2011

Contents

1 Introduction to the course 1

1.1 What this course is about . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Data representations 7

2.1 Bits, nibbles, bytes and words . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 4-bit Binary, Hexadecimal and Decimal Digits . . . . . . . . . . . . . . . . . 9

2.3 Binary arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.4 Converting from binary to decimal . . . . . . . . . . . . . . . . . . . . . . . 10

2.4.1 Whole numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.4.2 Real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.5 Binary integer arithmetic on computers . . . . . . . . . . . . . . . . . . . . . 11

2.5.1 Two’s-complement integer arithmetic . . . . . . . . . . . . . . . . . . 12

2.6 32-bit Binary, Hexadecimal, Unsigned and Signed Integers . . . . . . . . . . 13

2.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Algorithms and Pseudocodes 23

3.1 What is an algorithm? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2 Flowchart representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.3 Pseudocode representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.4 Decisions, conditionals, branch instructions . . . . . . . . . . . . . . . . . . 25

3.5 Looping or jump instructions . . . . . . . . . . . . . . . . . . . . . . . . . . 25

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3.6 Sequencing, branching and looping . . . . . . . . . . . . . . . . . . . . . . . 26

3.7 A mini-summary before the examples . . . . . . . . . . . . . . . . . . . . . . 27

3.8 Some examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.8.1 The Breakfast algorithm, or, Bielajew’s Sunday Morning Internationally-famous pancakes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.8.2 Solve for x: Ax2 +Bx+ C = 0 where A,B,C are arbitrary constants 29

3.8.3 Iteration: A summing loop . . . . . . . . . . . . . . . . . . . . . . . . 33

3.8.4 Iteration: A product loop . . . . . . . . . . . . . . . . . . . . . . . . 36

3.9 An aside on computer architecture . . . . . . . . . . . . . . . . . . . . . . . 40

3.9.1 What does S = S + 1 mean? . . . . . . . . . . . . . . . . . . . . . . . 40

3.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.11 Projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4 Getting started in C++ 49

4.1 Simple input/output (I/O): A first program in C++ . . . . . . . . . . . . . . 49

4.2 Compiling, linking, loading and running . . . . . . . . . . . . . . . . . . . . 52

4.3 Declaring and initializing variables . . . . . . . . . . . . . . . . . . . . . . . 55

4.4 Integer math in C++ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.5 Floating point math in C++ . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

4.6 The if/else if/else construct . . . . . . . . . . . . . . . . . . . . . . . . . 60

4.7 Logical expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.7.1 Logical expressions with AND or OR . . . . . . . . . . . . . . . . . . 66

4.7.2 Mixed arithmetic and logical expressions . . . . . . . . . . . . . . . . 67

4.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

4.9 Projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

5 Loops 83

5.1 The while loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

5.2 The do/while loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

5.3 The for loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

CONTENTS v

5.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

5.5 Projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

6 Early Abstraction—Functions 123

6.1 Motivation for functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

6.2 User-defined functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

6.3 Example: A problem tackled with teamwork . . . . . . . . . . . . . . . . . . 134

6.4 Call by value, call by reference, reference parameters . . . . . . . . . . . . . 138

6.4.1 The address of a variable . . . . . . . . . . . . . . . . . . . . . . . . . 138

6.4.2 Call-by-value vs. call-by-reference . . . . . . . . . . . . . . . . . . . . 140

6.5 The rules of scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

6.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

6.7 Projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

7 More Variable Types, Data Abstraction 167

7.1 Representation of floating-point numbers . . . . . . . . . . . . . . . . . . . . 167

7.1.1 When is one not one? Floating point anomalies. . . . . . . . . . . . . 174

7.2 char: the character variable . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

7.2.1 Character strings; the string class . . . . . . . . . . . . . . . . . . . . 177

7.3 The vector class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

7.3.1 Introduction to vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 179

7.3.2 Declaring and using vectors . . . . . . . . . . . . . . . . . . . . . . . 180

7.3.3 Vector syntax and rules: . . . . . . . . . . . . . . . . . . . . . . . . . 182

7.3.4 Vectors and functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

7.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

8 More Data Abstraction, Arrays, Structures 199

8.1 Arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

8.1.1 Declaring a 2-dimensional array . . . . . . . . . . . . . . . . . . . . . 200

8.1.2 Initializing a 2-dimensional array . . . . . . . . . . . . . . . . . . . . 201

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8.1.3 Passing a two-dimensional array to a function . . . . . . . . . . . . . 204

8.2 Character arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

8.2.1 One dimensional character arrays . . . . . . . . . . . . . . . . . . . . 209

8.2.2 Two dimensional character arrays . . . . . . . . . . . . . . . . . . . . 211

8.3 Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

8.3.1 The structure definition . . . . . . . . . . . . . . . . . . . . . . . . . 212

8.3.2 Where can structures be defined? . . . . . . . . . . . . . . . . . . . . 213

8.3.3 The members of a structure . . . . . . . . . . . . . . . . . . . . . . . 213

8.3.4 Declaring structure variables . . . . . . . . . . . . . . . . . . . . . . . 213

8.3.5 Assigning values to structure members . . . . . . . . . . . . . . . . . 213

8.3.6 Re-assigning values of structure members . . . . . . . . . . . . . . . . 214

8.3.7 Function call-by-value of a structure . . . . . . . . . . . . . . . . . . 214

8.3.8 Function call-by-reference to a structure . . . . . . . . . . . . . . . . 215

8.3.9 Structure arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

8.3.10 Example using structures and arrays: Ion transport . . . . . . . . . . 218

8.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222

8.5 Projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228

9 Miscellaneous Topics 239

9.1 Generating random numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

9.1.1 Example: Integrating functions by random sampling . . . . . . . . . 246

9.2 Simple Sorting: The bubble or sinking sort . . . . . . . . . . . . . . . . . . . 248

9.3 Recursion—A function calling itself . . . . . . . . . . . . . . . . . . . . . . . 249

9.4 Input and Output using files . . . . . . . . . . . . . . . . . . . . . . . . . . . 252

9.4.1 File and stream handling in C++ . . . . . . . . . . . . . . . . . . . . 252

9.4.2 Creating sequential access files and writing to them . . . . . . . . . . 255

9.4.3 Reading from sequential access files . . . . . . . . . . . . . . . . . . . 258

9.5 Command line arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

9.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264

CONTENTS vii

10 A Potpourri of Applications 267

10.1 Finding a zero using the “binary chop” . . . . . . . . . . . . . . . . . . . . . 267

11 Programming in MATLAB 271

11.1 A taste of MATLAB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271

11.2 Arrays in Matlab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273

11.3 Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

11.3.1 The for loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

11.3.2 The while loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282

11.4 The if/else construct . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282

11.5 Some Matlab terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283

11.6 Operators in MATLAB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

11.6.1 Math operators in Matlab . . . . . . . . . . . . . . . . . . . . . . . . 284

11.7 Pointwise operators in MATLAB . . . . . . . . . . . . . . . . . . . . . . . . 285

11.7.1 Logical Operators in Matlab . . . . . . . . . . . . . . . . . . . . . . . 286

11.8 M-files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288

11.8.1 Script M-files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288

11.8.2 Function M-files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292

11.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295

11.10Projects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327

11.10.1Supplementary material: Trajectories without air resistance . . . . . 346

11.10.2Supplementary material: Trajectories with air resistance . . . . . . . 346

11.10.3Supplementary material: Stepping algorithms . . . . . . . . . . . . . 347

11.10.4Supplementary material: Trajectories of objects in the universe . . . 354

11.10.5Supplementary material: Stepping algorithms . . . . . . . . . . . . . 355

12 Graphics 367

12.1 Two Dimensional Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367

12.1.1 A basic plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367

12.1.2 Using different line colors . . . . . . . . . . . . . . . . . . . . . . . . 367

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12.1.3 Making titles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370

12.1.4 Axis labels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370

12.1.5 Different line styles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370

12.1.6 Point plots, plotting with point symbols . . . . . . . . . . . . . . . . 372

12.1.7 Plotting more than one thing at once . . . . . . . . . . . . . . . . . . 377

12.1.8 Subplots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378

12.2 Three dimensional graphics . . . . . . . . . . . . . . . . . . . . . . . . . . . 379

12.2.1 Plot3 plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379

12.2.2 Mesh plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379

12.2.3 Axis labelling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381

12.2.4 meshc and meshz plots . . . . . . . . . . . . . . . . . . . . . . . . . . 382

12.2.5 Surface plots using surf . . . . . . . . . . . . . . . . . . . . . . . . . . 382

12.2.6 Surface plots using surfc . . . . . . . . . . . . . . . . . . . . . . . . . 383

12.2.7 Surface plots using surfl . . . . . . . . . . . . . . . . . . . . . . . . . 383

12.2.8 Contour plots using contour . . . . . . . . . . . . . . . . . . . . . . . 384

12.2.9 Contour plots using contour3 . . . . . . . . . . . . . . . . . . . . . . 384

12.2.10Contour plots using pcolor . . . . . . . . . . . . . . . . . . . . . . . . 384

12.2.11Contour plots using contourf . . . . . . . . . . . . . . . . . . . . . . . 385

12.2.12Contour plots with labels . . . . . . . . . . . . . . . . . . . . . . . . . 385

13 Miscellaneous topics 387

13.1 Pitfall review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387

13.1.1 Review of for loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387

13.1.2 Review of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390

13.2 The MATLAB way . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395

13.3 Selected Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395

13.3.1 Free fall with air resistance . . . . . . . . . . . . . . . . . . . . . . . . 395

13.3.2 Diffusion of charged ions in an electric field . . . . . . . . . . . . . . . 400

13.3.3 Calculation of electric fields . . . . . . . . . . . . . . . . . . . . . . . 402

CONTENTS ix

13.4 Summary of the Course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404

14 Programming Style Guide for C++ 405

15 Syntax reference for beginning C++ 413

16 Syntax reference for more advanced C++ 417

17 Syntax reference for MATLAB 421

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Chapter 1

Introduction to the course

Computers are stupid. They only give answers.

Computers are blindly faithful. They only do what you tell them to do.

Computers are dumb. They do not do what you want them to do.

1.1 What this course is about

This course is called “ENG101: Introduction to Computers and Programming”.

The College of Engineering Bulletin describes it as follows (emphasis is mine):

Algorithms and programming in C++ and MATLAB, computing as a tool in engineering,introduction to the organization of digital computers.

Let’s define a few things first...

What is an engineer? What is engineering?

The American Heritage Dictionary defines engineering as The application of science to prac-tical ends, such as the design, manufacture, and operation of structures, machines, andsystems. An engineer is defined as someone who practices engineering as a profession.

Our definition is somewhat broader. Engineering is a broad range of technical endeavorthat starts with the basic sciences (trying to characterize and understand the basic laws ofnature), integrates and systematizes this knowledge, builds things based on these principles,manufactures these “things” on a large scale, and makes them available for societal benefit.If we broaden our definition of “things” to ideas, methods, and algorithms (a prescriptionfor doing something, like a recipe in cooking), as well as physical devices, then that capturesthe essence of what engineering is today.

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2 CHAPTER 1. INTRODUCTION TO THE COURSE

In our College of Engineering you will encounter faculty and staff who might otherwise havebeen called Mathematicians, Physicists, Chemists and Biologists.

What is a computer?

Again, we appeal at first to some standard reference texts.

American Heritage Dictionary1. One (a person) who computes.Some of the best examples:- Richard Feynman’s computers—teams of people working out calculations during theManhattan Project

- Calculation of ballistics tables for WWII- Calculation of sine/cosine/logarithm tables2. A device (a machine) that computes, especially a programmable electronic machinethat performs high-speed mathematical or logical operations or that assembles, stores,correlates, or otherwise processes information.

Oxford English Dictionary1. One who computes; a calculator, reckoner; spec. a person employed to make calcu-lations in an observatory, in surveying, etc. (Earliest reference 1646).2. A calculating-machine; esp. an automatic electronic device for performing mathe-matical or logical operations; freq. with defining word prefixed, as analogue, digital,electronic computer (see these words). (Earliest reference 1897).

McGraw-Hill Encyclopedia of Science and TechnologyA device that receives, processes, and presents information. The two basic types ofcomputers are analog and digital. Although generally not regarded as such, the mostprevalent computer is the simple mechanical analog computer, in which gears, levers,ratchets, and pawls perform mathematical operations – for example, the speedometerand the watt-hour meter (used to measure accumulated electrical usage). The generalpublic has become much more aware of the digital computer with the rapid proliferationof the hand-held calculator and a large variety of intelligent devices, ranging fromtypewriters to washing machines.

Our definition A “device” that

1. accepts some form of input (usually disorganized or of some highly specific nature),

2. responds to this input information in a “pre-determined” (programmed) way,

3. produces some form of output (usually more organized or coherent than the input)

Some common examples of computers

1.1. WHAT THIS COURSE IS ABOUT 3

• Personal: Wristwatch, electronic car keys, phone, radio, GPS device, calculator, lap-top...

• In the home: Stove, refrigerator, freezer, microwave oven, clocks and timers, toaster,coffee machine, thermostat, washer and dryer, TVs, DVD/Blu-Ray players, audioequipment ... the list is endless.

• Automotive: fuel delivery, air/fuel mixture, spark advance, brakes (ABS), traction con-trol, suspension adjustment, climate control, navigation, parking assist, lane variationwarning, proximity warnings, computer-controlled driving (coming soon)...

• Mechanical analog computers: Wind-up clocks, toys, mechanical calculators...

• Electrical analog computers: Integrators, differentiators, differential equation solvers,Christmas light flashers (cheap ones!)...

• Digital computers: Calculators, PDAs, Palm computers, laptops, desktops, mini-computers, mainframe computers....

Digital computers?

Digital computers are the simplest of all! They only understand two things, which welabel “1” and “0”, or “on” or “off”, but more precisely, states of “high” or “low” voltage.Computers switch between these two states very, VERY quickly. Modern “run-of-the-mill”digital computers can throw about 100 billion switches per second! That’s enough to writeabout 10 million pages of text in one second.

• The input is rarely a collection of 0s and 1s, 011110000110... It can be in the form oftext that is converted to 0s and 1s or voltage that is “digitized” with an “analog-to-digital” converter. (Musical recording, for example.)

• The output is similarly rarely a collection of 0s and 1s. It can be 1D data (words,numbers), 2D data (functions, charts, pictures, music), 3D data (video, surfaces, time-evolving functions)...

• The processing or computing or programming part is what the bulk of the courseis concerned with. The programming responds to the input in some pre-determinedway. In some digital computers the programming is fixed, for example, a digital clockor a digital thermometer. In the most interesting cases, the program can be changed.In this course we will learn to program programmable digital computers.


How are digital computers programmed?

You can either start off the computer with the 0s and 1s in the right place, or, you use a pro-gramming language. Programming languages provide the prescription, the pre-determinedsteps that instruct the computer how to respond to the various input it gets. We will learntwo languages in this course.

C++ is a very popular (one of the most popular and common) language, a general-purposelanguage that can be used for all aspects of computing. It is the best language to useif one wants to teach many of the basic operations that a computer can perform.

MATLAB is a language that is suited for scientific and engineering applications, designedfor getting answers quickly, with the minimum of effort, with reasonable guaranteethat the answer is correct. It also provides graphical output with a minimum of effort.

What this course is REALLY about?

The hidden agenda

This course is all about putting ideas into action. It is about taking understood notions ofhow things work and building something, developing a numerical solution for some specificproblem. This is what engineering is all about. The course will try to put your knowledgeof mathematics and physics to practical use. Some of the problems we may solve in thiscourse...

• How can I program a computer to play a simple game?

• How far can a golf ball fly into a 50 mile/hour wind?

• How can one find the shortest distance between two points when the landscape is notflat?

• If I take 10,000 steps in random directions, how far, on average, will I be from whereI started?

• How can I do calculus on a computer (integrate, differentiate, find maxima and min-ima)?

• Where should I place a radiator in my room so that I am warm and do not have topay too much for the heat?

• How can I make a vacuum?

1.1. WHAT THIS COURSE IS ABOUT 5

What this course is NOT REALLY about?

The ugly stuff

• It is not primarily a course on computer language syntax. It may seem like it at times,but it really isn’t. Learning syntax is just a means to an end.

• It is not really a course on operating systems, computer hardware, or the fine detail ofhow computers work. We will discuss this but only enough to get the job done.

Educational Objectives

Engineering 101, Introduction to Computers and Programming, is intended to give you anexposure to, and a chance to practice, algorithmic thinking. It does so by asking you toactually solve problems using algorithms. This may be different from other courses that youhave taken, because we do not tell you exactly how to solve the assigned problems. Rather,we provide you with tools that will allow you to solve them and some examples of howproblems are solved. But you must creatively assemble the solution yourself.

Note that neither C++ nor MATLAB languages per se are the keys to the course. Weteach you these languages so that you will have a means to express algorithms for executionon a computer, but a great many other computer languages could have been selected tosupport our central purpose of getting you to think about solving problems through a setof procedural steps. Because programming languages such as C++ and MATLAB containmany constructs and demand great attention to detail, it is easy to lose sight of the trueintellectual center of the course. But keeping a focus on algorithmic thinking will makewriting programs less difficult and less time consuming.

The educational objectives of Engineering 101 include:

• To introduce students to algorithmic thinking in their approach to problem solving.

• To teach students to implement algorithms in ANSI C++.

• To teach students to implement algorithms in MATLAB.

• To have students apply their knowledge of elementary physics and calculus.

• To provide students with a foundation on which to base their later applications ofcomputers in engineering.

Students who successfully complete Engineering 101 (receive a grade of C or better) should:

• Design algorithms to accomplish clearly specified tasks.


• Display knowledge of ANSI C++ syntax and semantics.

• Display knowledge of MATLAB syntax and semantics.

• Successfully implement clearly stated algorithms in C++.

• Successfully implement clearly stated algorithms in MATLAB.

• Be prepared to engage in discipline-specific instruction in computing.

1.2 Problems

1. In the broad definition of “computer” in this chapter, how many computers can youfind: On yourself? In your room? In a car? In your home?

2. A typical 3rd grader is a lot smarter than any computer. They can read, do simplemath, play a musical instrument, memorize poetry and write stories. Generally, withsufficient motivation, they will also do as they are told. How many ways could youinstruct a 3rd grader to determine the value of π without the aid of a computer orcalculating device?

Chapter 2

Data representations

2.1 Bits, nibbles, bytes and words

bit The digital computer processing hardware is made up of transistors, each of which canexist in only one of two possible states, a “state” of “high voltage” or “low voltage”.This state of high/low (or on/off) is represented abstractly as 1/0 (one/zero). A “bit”is some piece of microscopic (usually!) hardware that can exist in either of these twostates as determined by a programmer or computer designer. In addition to the voltagestates in a transistor in a computer’s memory chips or processing chips, there are manyothers forms: magnetic domains in a floppy disk, hard disk or magnetic tape, “pits”in a CD or DVD, holes in paper tape or computer cards. Nowadays, there are billionsor trillions of bits of various forms in a typical computer. A bit may be simple, butit is usually so small, typically well below a micrometer across (1 micrometer = 10−6

m), that we can conceive of cramming billions of them into a relatively small volumeof space. Some bits, particularly those in the processing chips of a computer, can alsobe made to change states very rapidly, well under one nanosecond. (1 nanosecond =10−9 s.) So, we can do a lot of different things with lots of bits in a very short span oftime.

nibble A “nibble” is a collection of 4 bits. A nibble can be in one of 24 = 16 separatecombinations of bits.

byte A “byte” is a collection of 2 nibbles, or 8 bits. A byte can be in one of 28 = 256separate combinations of bits. This is enough to, for example, associate with eachunique pattern one of the characters of the English alphabet (upper and lower case)plus all the common special characters we may use, for example, in writing Englishprose or a technical document, e.g. ‘!$&(),:;“’?. And, there would be lots of room leftover to make other interesting associations.

7

8 CHAPTER 2. DATA REPRESENTATIONS

word A “word” is a collection of bytes—typically 4 for 232 = 4, 294, 672, 296 possible com-binations. We can do a lot with all the different representations of “words”. We willsee, later in the lecture, how whole numbers can be associated with the various wordpatterns.

The notion that many bits can conspire, in very large numbers, to produce wonderfullyintricate objects is illustrated in the following pictures of my dog, Matilda Skip Bielajew II.(Yes, that is her real name!) The first picture was generated from a file that was about 12Megabytes, containing 96 million bits. It looks like real life. Yet, even when we zoom a little,we start to see the effects of the underlying bit structure. Look at how her whiskers appearjagged. Under extreme magnification (one of her whiskers, again) we see that the picture ismade up of flat, uniformly colored “tiles”. Each one of these tiles is represented by 24 bits,8 bits each for red, green and blue intensity, that, to all appearances, affords an apparentlyuniform transition of color from tile to tile.

2.2. 4-BIT BINARY, HEXADECIMAL AND DECIMAL DIGITS 9

2.2 4-bit Binary, Hexadecimal and Decimal Digits

4-bits of data (a “nibble”) can represent or be represented by several different things. It iscommon for computer programmers to use “hexadecimal” notation (0, 1, 2...9, A, B...F) asa “shorthand” for (0000, 0001, 0010...1001, 1010, 1011...1111). See the following table.

However, collection of a certain number of bits has to be given an “interpretation”—one thatis sensible to humans. One such assignment (which is completely arbitrary and is decidedupon by the computer programmer or designer) is to assign “decimal” numbers to the stringsof bits. Such an assignment could look like:

Binary Hexadecimal Decimalrepresentation representation assignment

0000 0 00001 1 10010 2 20011 3 30100 4 40101 5 50110 6 60111 7 71000 8 81001 9 91010 A 101011 B 111100 C 121101 D 131110 E 141111 F 15

We will use this particular representation in doing some simple arithmetic in the exercisesand the assignments.

2.3 Binary arithmetic

Let us pretend that we have to do some arithmetic (additions and multiplications) using onlybinary numbers. The numbers 2,3,....9 do not exist! Only 0 and 1 exist. We also assumethat the operations of addition and multiplication are commutative (i.e. independent of theorder of the mathematical operation).

The addition of “fictional” binary numbers–base-2 arithmetic looks like:


0 + ? = ? + 0 = ?1 + 1 = 1010 + 1 = 1 + 10 = 1111 + 1 = 1 + 11 = 10010 + 10 = 10 + 10 = 100..1111 + 111 = 111 + 1111 = 10110...

The multiplication of “fictional” binary numbers–base-2 arithmetic looks like:0 × ? = ? × 0 = 01 × ? = ? × 1 = ?10 × 10 = 10 × 10 = 10010 × 11 = 11 × 10 = 11011 × 11 = 1001...1111 × 111 = 111 × 1111 = 1101001...

The arithmetic procedure is analogous to the more familiar base-10 system (that arose simplybecause of our anatomy—10 fingers). Additions of columns of numbers, subtraction, mul-tiplication, the procedure of long division, can be accomplished by analogous procedures.There is no limit to how long a “fictional” base-2 number can be and we may have to carryalong some extra information with each number—its sign.

2.4 Converting from binary to decimal

2.4.1 Whole numbers

Humans are usually quite awful at doing binary arithmetic. If you are doing such a calcula-tion, you can check by converting the binary numbers into their decimal equivalents.

In general, to convert a whole binary number to a whole decimal number, consider a binarynumber of the form:

bnbn−1bn−2 · · · b2b1b0,

2.5. BINARY INTEGER ARITHMETIC ON COMPUTERS 11

where the bi’s are either 0 or 1. We compute a decimal number using the following formula:

d10 = bn × 2n + bn−1 × 2n−1 + bn−2 × 2n−2 · · · b2 × 22 + b1 × 21 + b0 × 20.

For example, the result of the two more difficult calculations above can be checked as follows:

101102 = 1× 24 + 0× 23 + 1× 22 + 1× 21 + 0× 20 = 24 + 22 + 21 = 1610 + 410 + 210 = 2210

is the result of one of the additions [11112(1510) + 1112(710)] above, while

11010012 = 26 + 25 + 23 + 20 = 6410 + 3210 + 810 + 110 = 10510

is the result of one of the multiplications [11112(1510) × 1112(710)] above. Note that thesubscript “2” indicates a binary or base-2 number while the subscript “10” indicates adecimal number.

2.4.2 Real numbers

To convert a real(non-whole) binary number to a real decimal number, consider a binarynumber of the form:

bnbn−1bn−2 · · · b2b1b0 . b−1b−2 · · · b−(m−2)b−(m−1)b−m,

where the bi’s are either 0 or 1 and a period, “.”, separates the whole part, bi where i ≥ 0from the fractional part, bi where i < 0. We compute a real decimal number using thefollowing formula:

r10 = bn×2n+bn−1×2n−1 · · · b1×21+b0×20+b−1×2−1+b−2×2−2 · · ·+b−(m−1)×2−(m−1)+b−(m)×2−m.

It is not a whole lot different from the decimal numbers you are used to. However, sometimesthe formulae can look a little messy.

2.5 Binary integer arithmetic on computers

Computers, because of their comprehension of data in binary terms, naturally do theirarithmetic using the base-2 representation. The circuitry within a computer is most easilyset up this way. However, when it comes to base-2 arithmetic and computers, there is asubtle difference! For practical reasons, computers must deal with integers that are finitein length. 16 bits, representing 65,536 separate possibilities, used to be the norm. Today,


32 bits is common (4294967296 possibilities). Soon, 64 bits (about 18×1018) will be thestandard. Some computers nowadays permit hardware 64-bit integer arithmetic.

Let us consider 32-bit computers only—the most common type of hardware available today.In the course of doing an arithmetic calculation, any bits that are set beyond the 32-bitboundary are lost—they fall into the “bit bucket” and disappear.

2.5.1 Two’s-complement integer arithmetic

A string of 32-bits can represent many things but we will focus on hexadecimal, unsignedand signed integers. These signed integers must also carry information as to the sign of thenumber. Historically, there have been several ways to do this. The “industry” has “settled”on “two’s-complement arithmetic” because it simplifies the process of adding positive, nega-tive or mixed numbers. In this scheme, to negate a signed integer, one takes the complement(change all the 1s to 0s and all the 0s to 1s) of the binary representation and adds 1 to it.This is seen in the table at the end of this chapter.

Let’s give the notation i to the complement of the integer i. So, for any i,i+ i = 11111111111111111111111111111111. That is:

any 32-bit pattern+ complement of the above= 11111111111111111111111111111111

and

11111111111111111111111111111111+ 00000000000000000000000000000001= 00000000000000000000000000000000

since the 33rd bit can not be set! In two’s-complement arithmetic the expression i − j iscalculated as i+ j+1. So you see, computers can not really add and subtract, they can onlyadd (and take complements)!

A few examples help illustrate the simplicity of the two’s-complement scheme. Consider a4-bit representation of the number 310 = 00112 and 410 = 01002. There are two tricks youshould apply here: a) every time you see a negative sign in front of a bit pattern, i, to be usedin a calculation, convert it to a positive sign, change i to i+ 1 and do the binary addition,and, b) if a final result has a leading order (leftmost) bit set to 1, it is a negative number,so, extract the minus sign and convert the resultant bit pattern j to j + 1.

310 + 410 = 00112 + 01002 = 01112 = 710

310 − 410 = 00112 − 01002 = 00112 + 11002 = 11112 = −00012 = −110−310 − 410 = −00112 − 01002 = 11012 + 11002 = 10012 = −01112 = −710

2.6. 32-BIT BINARY, HEXADECIMAL, UNSIGNED AND SIGNED INTEGERS 13

See how easily it all worked out! This is why the two’s-complement approach has beenadopted. In order to handle negative integers one can convert them to their two’s-complementcounterparts and add. Otherwise, special circuitry would have to be developed for handlingnegative numbers and that would be less efficient.

2.6 32-bit Binary, Hexadecimal, Unsigned and Signed

Integers

32-bit computers can represent 232 = 4, 294, 672, 296 different things but ONLY 4, 294, 672, 296different things. Now we consider the conventional assignments to “unsigned” and “signed”integers. There is also the hexadecimal representation that is associated with every 4-bitsequence, as indicated in the table below.

The unsigned integers associate the decimal number 0 to a string of 32 0 bits. The rest areformed by the addition by 1 in either binary or decimal. The maximum unsigned integer,therefore, is 4, 294, 672, 295. Adding a one to this causes all the 32 bits to go back to zero,exactly as if you had “cycled” the odometer in your car! It’s the same principle. There isno 33rd bit, so you can imagine it just falls off the end.

The signed integers have the same sequence as the unsigned integers up to 231 − 1 =2147483647 which is the bit pattern 01111111111111111111111111111111. Adding a oneto this gives a bit pattern of 10000000000000000000000000000000 which is interpreted as−231 = −2147483648. Note that these two are complements of each other and that they addup to11111111111111111111111111111111, which has the interpretation “-1”, consistent with the“two’s-complement” scheme described above.


Binary Hexadecimal Unsigned Int Signed Int00000000000000000000000000000000 00000000 0 000000000000000000000000000000001 00000001 1 100000000000000000000000000000010 00000002 2 200000000000000000000000000000011 00000003 3 300000000000000000000000000000100 00000004 4 4

. . . .

. . . .

. . . .00000000000000000000000000001111 0000000F 15 1500000000000000000000000000010000 00000010 16 1600000000000000000000000000010001 00000011 17 17

. . . .

. . . .

. . . .01111111111111111111111111111110 7FFFFFFE 2147483646 214748364601111111111111111111111111111111 7FFFFFFF 2147483647 214748364710000000000000000000000000000000 80000000 2147483648 -214748364810000000000000000000000000000001 80000001 2147483649 -2147483647

. . . .

. . . .

. . . .11111111111111111111111111111100 FFFFFFFC 4294967292 -411111111111111111111111111111101 FFFFFFFD 4294967293 -311111111111111111111111111111110 FFFFFFFE 4294967294 -211111111111111111111111111111111 FFFFFFFF 4294967295 -1

2.7. PROBLEMS 15

2.7 Problems

1. octal + octal = octal

Note: An “octal” is a number represented by only 3 bits.

Consider a fictitious computer, one that can only do 3-bit binary arithmetic. (Anythingpast the 3’rd bit can not be represented. See the notes for Lecture 2 for some relatedexamples.) Complete the following addition table, that is, add the octal in i’th row tothe octal in the j’th column and put the resultant octal in the box located at the i’throw and j’th column. There are 3 examples worked out for you. For these examples,show your calculations in long-hand.

000 001 010 011 100 101 110 111000001010011100101 010110111 011 110

2. octal × octal = octal

This is a similar exercise to the above except that you are to multiply the octals. Besure to show your work for the 3 examples given.

000 001 010 011 100 101 110 111000001010011100101 001110111 100 001


3. Bits, nibbles and integers. Is it a hex?

Your imaginary computer represents integers using 4 and only 4 bits! The result of anarithmetic operation with two integers always results in a 4-bit result. Higher-orderbits are “lost”.

The following table shows the binary, hexadecimal, unsigned integer, and signed inte-ger representations for all the possible bit patterns that can be represented by yourcomputer.

Binary Hexadecimal Unsigned Int Signed Int0000 0 0 00001 1 1 10010 2 2 20011 3 3 30100 4 4 40101 5 5 50110 6 6 60111 7 7 71000 8 8 -81001 9 9 -71010 a or A 10 -61011 b or B 11 -51100 c or C 12 -41101 d or D 13 -31110 e or E 14 -21111 f or F 15 -1

Do your calculations in binary arithmetic and SHOW YOUR WORK! Quote yourresults of your calculations in binary, integer, unsigned and hexadecimal integer repre-sentations.

(a) What is the sum of the integers 2 and 3?

(b) What is the sum of the integers -2 and -3?

(c) What is the sum of the integers -8 and -8?

(d) What is the product of the integers 2 and 3?

(e) What is the product of the integers -2 and -3?

(f) What is the product of the integers -8 and -8?

2.7. PROBLEMS 17

4. The interbreeding hex: hex + hex = hex

Consider a fictitious computer, one that can only do 4-bit binary arithmetic. (Anythingpast the 4’th bit can not be represented. See the notes for Lecture 2 for some relatedexamples.) Complete the following addition table, that is, add the hex in i’th row tothe hex in the j’th column and put the resultant hex in the box located at the i’throw and j’th column. Some of the answers are given to you. Fill in the empty boxes.Remember that addition commutes. Therefore the actual work is halved and you neednot do the boxes that are filled with X’s.

+ 0 1 2 3 4 5 6 7 8 9 A B C D E F0 0 X X X X X X X X X X X X X X X1 1 2 X X X X X X X X X X X X X X2 2 3 4 X X X X X X X X X X X X X3 3 4 6 X X X X X X X X X X X X4 4 5 8 X X X X X X X X X X X5 5 6 a X X X X X X X X X X6 6 7 c X X X X X X X X X7 7 8 e X X X X X X X X8 8 9 0 X X X X X X X9 9 a 2 X X X X X XA a b 4 X X X X XB b c 6 X X X XC c d 8 X X XD d e a X XE e f c XF f 0 e


5. Oh, hex, won’t it ever stop: hex × hex = hex

This is a similar exercise to the previous one except that you are to multiply the hexes.

X 0 1 2 3 4 5 6 7 8 9 A B C D E F0 0 X X X X X X X X X X X X X X X1 0 1 X X X X X X X X X X X X X X2 0 2 4 X X X X X X X X X X X X X3 0 3 9 X X X X X X X X X X X X4 0 4 0 X X X X X X X X X X X5 0 5 9 X X X X X X X X X X6 0 6 4 X X X X X X X X X7 0 7 1 X X X X X X X X8 0 8 0 X X X X X X X9 0 9 1 X X X X X XA 0 a 4 X X X X XB 0 b 9 X X X XC 0 c 0 X X XD 0 d 9 X XE 0 e 4 XF 0 f 1

2.7. PROBLEMS 19

6. A fistful of (binary) digits

Your imaginary computer represents integers using 5 and only 5 bits! Fill in the blankspots in the following table. The signed integers follow the two’s-complement rulediscussed in class.

Binary Hexadecimal Unsigned Int Signed Int00000 00 0 000001 01 1 100010 02 2 200011 03 3 300100 04 4 400101 05 5 500110 06 6 600111 07 7 701000 08 801001 09 901010 0A 1001011 0B 1101100 0C 1201101 0D 1301110 0E 1401111 0F 1510000 10 1610001 11 1710010 1210011 1310100101011011010111110001100111010110111110011101 -311110 -211111 -1


7. These digits will give you the fidgets.

Your imaginary computer still represents integers using 5 and only 5 bits! The result ofan arithmetic operation with two integers always results in a 5-bit result. Higher-orderbits are “lost”.

Do your calculations in binary arithmetic and SHOW YOUR WORK! Quote yourresults of your calculations in binary, signed integer (2s-complement scheme), unsignedand hexadecimal integer representations. You may use the table that you completedin the previous question.

(a) What is the sum of the unsigned integers 3 and 4?

(b) What is the sum of the unsigned integers 16 and 17?

(c) What is the sum of the signed integers -2 and -3?

(d) What is the sum of the signed integers -16 and -16?

(e) What is the product of the unsigned integers 3 and 4?

(f) What is the product of the signed integers -2 and -3?

(g) What is the product of the signed integers -16 and -16?

2.7. PROBLEMS 21

8. Bits and bytes? Again!

In the following table, the column 1 represents bit patterns on a 6-bit computer.

(a) In the 2’nd column, write in the complement of the bit pattern in column 1

(b) In the 3’rd column, write in the two’s complement of the bit pattern in column 1

(c) In the 4’th column, write in the two’s complement of the bit pattern in column 3

(d) In the 5’th column, put a check mark if the bit pattern in column 1 represents apositive signed int.

(e) In the 6’th column, put a check mark if the bit pattern in column 1 represents anodd unsigned int.

(f) In the 7’th column, put a check mark if the bit pattern in column 1 represents anunsigned int that is divisible by 2.

(g) In the 8’th column, put a check mark if the bit pattern in column 1 represents anunsigned int that is divisible by 4.

(h) If the bit pattern in column 1 represents an odd unsigned int, write its unsignedint representation in column 9.

Column 1 Column 2 Column 3 Column 4 Col 5 Col 6 Col 7 Col 8 Column 9

000010111001110110000001001101110010010010110110101111101010


9. A bit challenging

In the following table:1) Convert the first 5 32-bit bit patterns to hex2) indicate which bit patterns are divisible by 2 (/2)3) indicate which bit patterns are divisible by 4 (/4)4) indicate which bit patterns represent positive or negative integers in two’s-complementrepresentation (+/-?)5) indicate which bit pattern represents the largest integer in two’s-complement repre-sentation (>?)6) indicate which bit pattern represents the smallest integer in two’s-complement rep-resentation, negative with the greatest magnitude (<?)

32-bit bit patterns hex /2? /4? -/+? >? <?

1111 0001 0010 0110 1110 0101 0100 01101000 0010 0000 1100 1000 0010 1011 10111011 0001 1000 1011 0100 1111 0000 01110010 0001 1101 1001 0100 1101 0111 11100010 1000 1010 0001 0010 1010 0101 01111110 1100 1000 1000 0011 0101 1110 1101 X1111 0111 0000 1100 0011 1010 0010 1101 X0110 0111 0100 0000 0101 1000 1011 0000 X0011 1110 0000 0011 1100 0110 0000 1101 X0000 1110 1101 1001 1111 1011 0101 1000 X0110 1111 0001 1010 0010 1111 0110 1000 X1010 1000 0000 1110 1111 1010 0010 0011 X0000 0010 0011 0010 0100 1100 0001 0101 X0011 1100 1010 0111 1011 1011 0111 0000 X1100 0011 0100 1011 0110 1011 1111 0011 X1101 0001 1011 1110 1011 1010 0000 0001 X1101 0011 1110 1011 0100 0101 0100 1111 X1111 0011 0001 0111 1110 1110 0011 0000 X0100 1101 1110 1110 0000 0011 1011 1011 X0111 1011 1010 1111 0100 0010 0010 0101 X1101 1001 1101 1111 1101 1101 1010 1101 X1000 1101 1010 0101 0110 1010 1111 0110 X0100 0100 0110 1111 0001 0010 0111 0001 X0011 1110 1010 1100 1100 0111 0010 0111 X1110 1111 0111 1100 1001 0101 0001 1101 X1001 1000 1010 0101 1000 0010 1011 1010 X1011 1111 1000 0100 0011 0101 0011 0111 X1001 1010 0000 0011 1111 1010 0110 1111 9A03FA6F

Chapter 3

Algorithms and Pseudocodes

In this chapter we introduce algorithms and ways with which algorithms can be represented.At the end of the chapter, we also talk a little about computer architecture, as a way ofillustrating the operations that can happen inside a computer, when a part of an algorithmis executed.

3.1 What is an algorithm?

An algorithm is a set of well-defined, unambiguous steps or instructions to be carried out insequence to accomplish some task.

• An algorithm usually has a START point and a STOP point. Certainly any algorithmwe shall encounter in this book will. The START may require some inputs and theSTOP may provide some output.

• An algorithm is made of individual instructions.

• An instruction consists of a well-defined operation usually on some input (what theinstruction receives) and usually producing output (what the instruction produces).

• Each instruction is well-defined and its outcome predictable if the instruction operateson valid input.

• There is a direction of logic flow or sequencing. Once an instruction is executed, itpasses control to another instruction.

• There can only be a finite set of instructions.

• When executed (with valid input) an algorithm is guaranteed to terminate in a sensibleway.

23

24 CHAPTER 3. ALGORITHMS AND PSEUDOCODES

3.2 Flowchart representation

STEP 1

STEP 2

STEP 3

START

STOP

The above figure is a representation of a simple sequence of 3 steps with a beginning and anend. The START and STOP are represented by circles. A sequence step is represented bya rectangle. The logic flow is indicated by lines and arrows. This kind of representation iscalled a flowchart or schematic diagram. It looks a lot like a circuit diagram in electronics.

3.3 Pseudocode representation

An algorithm is a well-defined description of actions and their ordering. Pseudocode is an-other way of expressing the actions and ordering to be taken place on a computer in anexact way. Yet, pseudocode does it in an informal way, without getting bogged-down in thedetails or syntax of a particular computer language like C++, MATLAB or something else.Two programmers should be able to agree on pseudocode and write functionally equivalentcomputer programs—either in the same computer language or in different ones. Pseudocodecan be written in any human language in the form of an ordered (or numbered) list. Forexample, a pseudocode representation of the above flowchart would be:

START =⇒ Step 1 =⇒ Step 2 =⇒ Step 3 =⇒ STOP

A more conventional representation has the instruction flow (also called logic flow) goingfrom top to bottom:

3.4. DECISIONS, CONDITIONALS, BRANCH INSTRUCTIONS 25

STARTStep 1Step 2Step 3STOP

3.4 Decisions, conditionals, branch instructions

A “conditional” instruction or statement, also called a decision instruction, a branch or de-cision statement, is one in which a decision is made and the logic flow can follow differentpaths, depending on the answer. In pseudocode it can be represented as follows:

IF such-and-such THEN DOthis-and-that

OTHERWISE DOsomething-else

One could do the same thing with a flowchart, and it would look like:

YES NOSUCH&

SUCH

THIS&

THAT

SOMETHINGELSE

Note that we have introduced a new shape to handle the case where, depending on the input,a decision is made and the logic flow can follow two different paths. This is represented by a“diamond shape”. A branch statement or a decision statement is characterized by only onepath in and always two paths out.

3.5 Looping or jump instructions

A final ingredient is needed before we complete our abstract discussion of algorithms —theability for a statement to transfer control not to the next statement, but somewhere else in


the algorithm. This is called a “looping” or “jump” instruction.

Consider the following pseudocode:

ITEM 3..a set of other steps..

JUMP TO: ITEM 3

The stuff between ITEM 3 and JUMP TO: ITEM 3 inclusive, defines a loop. This isrepresented in a flowchart as follows with the direction of logic flow made clear with the linesand arrows.

ITEM 3

JUMP TOITEM 3

3.6 Sequencing, branching and looping

It can be shown that the 3 constructs of sequencing, branching and looping, are all that oneneeds to construct any computer algorithm! Thus, the “theory” of algorithmic flow boilsdown to understanding just these three things. There are many, many ways of expressingthese three things in a computer language like C++ or Matlab. But remember that it allboils down to just sequencing, branching and looping. These are the three pillars ofalgorithmic design!

Of course there is a lot of other stuff to be considered to make a computer language useful.We’ll see a lot of this in the course. However, as far as the flow of logic goes, we have doneit all at this point! You can now go and write an algorithm to do anything you want andexpress it in either pseudocode or flowchart form.

3.7. A MINI-SUMMARY BEFORE THE EXAMPLES 27

3.7 A mini-summary before the examples

PSEUDOCODE: The language and organization of pseudocode is not firmly established.Make any convention for yourself that makes your algorithm clear. Note the use of spacingand indenting, bold and italic fonts, all ways of emphasizing meaning.

FLOWCHARTS: It is also possible to write pseudocode diagrammatically in terms of flowcharts that represent the flow of computer instructions, or “logic flow”. Here there are someaccepted (but not rigorously adopted) standards. These charts look very much like circuitdiagrams. It is considered good practice to describe every variable you use and provide awell-defined starting point and ending point.

Pseudocode and flow charts are indispensable aids in getting a computer program “up andrunning” with a minimum of effort.

3.8 Some examples


3.8.1 The Breakfast algorithm, or, Bielajew’s Sunday MorningInternationally-famous pancakes

1. Start of The Breakfast Algorithm

2. If the day is not Sunday, have your usual boring bowl of Granola. For added excitement,slice in half a banana. Otherwise,

(a) In a big bowl, whisk together 1 and 1/2 cups white flour and 2 heaping teaspoonsof baking powder.

(b) In a separate bowl, beat two whole eggs (no shells please) together for one minuteor so.

(c) Add 1 cup of milk, preferably skimmed, to the eggs and mix.

(d) Add 1 tablespoon of Canola oil, 3 tablespoons of maple syrup, and one teaspoonof vanilla to the milk and eggs and mix.

(e) Add the wet stuff to the dry stuff.

(f) Take a big spoon and,

(g) Mix the stuff in the big bowl one whole turn.

(h) If there are lots of lumps in the mixture , go back to step 2.(g), otherwise go tostep 2.(i). Technical note: Don’t go overboard and make it completely smooth.Leave it a little lumpy for texture.

(i) Slice in a Granny Smith apple (no peel) and (optionally) a half cup of frozencranberries (well rinsed).

(j) Add a little bit of Canola oil to an electric fry pan preheated to 350◦F. Put injust enough to coat the bottom of the pan.

(k) Put the pancake dough into the pan, making approximately four 3–4” diametercircles. Turn the heat down to 300◦F. Fry for exactly 3 minutes with the pancovered, flip and fry for 3 more minutes with the pan uncovered.

(l) Take them out of the frying pan and place on a napkin to absorb excess oil.

(m) Eat with maple syrup and sliced strawberries on top. I DARE you to eat morethan 3! (It has never been done...)

3. Clean up your dishes.

4. End of The Breakfast Algorithm

This algorithm produces about 8 pancakes, enough for 4–6 people.

3.8. SOME EXAMPLES 29

3.8.2 Solve for x: Ax2 + Bx + C = 0 where A,B,C are arbitraryconstants

Definitions: x is an unknown “real” number, A,B,C are fixed constants that can take on anyvalid real value. We are trying to find the values of x for which Ax2 +Bx+C is zero. If weplot the function f(x) = Ax2+Bx+C, a parabola, it can look as shown in the figure below,for certain choices of A, B, and C. The x-axis can intersect the curve f(x) = Ax2+Bx+Conly 0, 1 or 2 times, depending on the values of A, B, and C.

NO SOLUTION

2 SOLUTIONS

Y

X

1 SOLUTION

Y

X

Y

X

First, we do the math in excruciatingly fine detail. It is essential to understand completelythe mathematics of what you are doing if you are to have a chance at writing a successfulalgorithm.

Ax2 +Bx+ C = 0

A(x2 +

B

Ax+

C

A

)= 0 N.B. A �= 0

x2 +B

Ax+

C

A= 0

x2 +B

Ax+

B2

4A2− B2

4A2+

C

A= 0


x2 +B

Ax+

B2

4A2=

B2

4A2− C

A

x2 +B

Ax+

B2

4A2=

B2 − 4AC

4A2(x+

B

2A

)2=

B2 − 4AC

(2A)2

x+B

2A= ±

√B2 − 4AC

2AN.B. B2 − 4AC ≥ 0

x = − B

2A±

√B2 − 4AC

2A

x =−B ±

√B2 − 4AC

2AA �= 0, B2 − 4AC ≥ 0

So, under some circumstances, (A �= 0, B2 − 4AC > 0), we have two solutions:

x =−B +

√B2 − 4AC

2Aand x =

−B −√B2 − 4AC

2A

corresponding to the last case in the preceding figure.

What if A �= 0, B2 − 4AC = 0? We have one solution:

x =−B

2A

corresponding to the middle case in the preceding figure.

What if B2−4AC < 0? No real solution. This corresponds to the first case in the precedingfigure.

What if A = 0? Re-analyze!

Bx+ C = 0 (y = Bx+ C Is an equation for a straight line!)

x =−C

BN.B. B �= 0

This is the place where a straight line crosses the x-axis.

What if A = 0 and B = 0? No solution for x! A straight line with zero slope does notintercept the x-axis. The equation is nonsense unless C = 0 as well. If A = B = C = 0 it iscorrect but not usually interesting!

Now we give the pseudo-code for this example, the solution to the quadratic equation.

START A,B,C are inputs, known to be real, fixed constants and x is the output, expectedto be real but unknown at the start


IF (B2 − 4AC < 0)PRINT (No solution because B2 − 4AC < 0)STOP

ELSE (i.e. B2 − 4AC ≥ 0)

IF (A �= 0)

IF (B2 − 4AC = 0)x = −B/(2A)PRINT (One solution: x, because B2 − 4AC = 0)STOP

ELSE (i.e. B2 − 4AC > 0)x1 = (−B +

√B2 − 4AC)/(2A)

x2 = (−B −√B2 − 4AC)/(2A)

PRINT (Two solutions are: x1 and x2)STOP

ELSE (i.e. A = 0)

IF (B �= 0)x = −C/BPRINT (Only one solution x, because A = 0)STOP

ELSE (i.e. B = 0)PRINT(No solution because A = 0 and B = 0)STOP

END OF ALGORITHM

Note:

• IF’s can be nested

• Use of indenting

• Multiple STOP-points!

• Not every piece of pseudocode is used every time

• If enough different A, B, C’s are tried, every piece of pseudocode will be used

• It will work for any value of A, B, C. The algorithm is robust

• Can represent in terms of a graphical flow chart

This is the flow chart for the quadratic formula:


A,B,C ARE FIXED X IS UNKNOWN

START

NOSOLUTION STOP

=/?

ISA 0

Y

0>B -4AC

NOSOLUTION STOP=/

X = -C / B

PRINT X

STOPPRINT X

STOP

X = -B / (2A)

B -4AC2= -B -X 2

B -4AC2= -B +X 1

2A

2A

X 2X 1PRINT ( , )

STOP

IS

?

N

Y

?

IS

N

-2

0B -4AC =

N

Y

B 0IS

?

2Y

N

Note that, although there is no reason to expect it, the flowchart designer expects thatthere will be two real solutions every time. This is reflected in the flowchart by this caserunning vertically down the page, with the exceptions hanging off to the right. This makesthe flowchart particularly easy to read.


3.8.3 Iteration: A summing loop

Before you read this section, if you have never had any experience with programming, thestatement S = S + 1 should justifiably seem to be completely wrong, since mathematicallyit is equivalent to saying 0 = 1! Just cancel the S’s from either side! So, if this is the case,kindly go and read the next section “An aside on computer architecture” and then comeback. Although mathematical statements and computer code statement may look the same,they are completely different. Mathematics expresses an idea, while a line of computer coderepresents an operation, or a set of individual operations.

Problem: Sum the positive integers 1, 2, 3 · · ·N up to N .

A dumb way that works:START N is an integer (fixed) and S is an integer (unknown)S = 1

S = S + 2

S = S + 3

.

.

.S = S + i (for the i’th statement)...S = S + N

PRINT(S)END OF ALGORITHM

This is the flow chart for the dumb way:


START

.

.

.

.

.

.

N FIXED NUMBERS IS UNKNOWN

S = 1

S = S + 2

S = S + N

S = S + i

S = S + 3

PRINT S

STOP

N statements—lots of repetition, very long if N is large.

A better way:

START N is an integer (fixed), S is an integer that will contain the sum . At the start, S isinitialized to 1, that is, its starting value will be 1. Another integer, i, is employed tocontain the value that is to be added to S. It is initialized to 1.

Start of counting loop:

• i = i + 1 (get the new value for i)

• S = S + i (add it to S)

• IF i = N jump to End of counting loop:


• Jump to the first statement in this loop

End of counting loop:PRINT(S)

END OF ALGORITHM

• Only a few “statements” to express the same idea

• Loops can be nested as well

• Can represent in terms of a graphical flow chart

This is the flow chart for the better way:

START

N FIXED NUMBERi = 1; S = 1

i = i + 1S = S + i

i = N ?

PRINT S

YES

NO

STOP

There is an even better way! There is a mathematical solution to this that was discoveredby Gauss (Karl Friedrich Gauss (1777–1855 AD)) as a schoolboy!

S =N(N + 1)

2


If N is even:

S = [1, 2, 3, ..., (N − 2), (N − 1), N ]

= [1, 2, 3, ..., (N/2)] + [(N/2 + 1), (N/2 + 2), (N/2 + 2), ..., N ]

= [1, 2, 3, ..., (N/2)] + [N, (N − 1), (N − 2), ..., (N/2 + 1)]

= [(N + 1), (N + 1), (N + 1), ..., (N + 1)] (N/2 terms)

=N

2(N + 1)

quod erat demonstratum

For odd N the proof is left to the reader.

3.8.4 Iteration: A product loop

Problem: Compute P = kN .

A very, VERY, dumb way (but one that works):

START k is an integer or a real (fixed), N is an integer (fixed) and P is an integer or a realnumber (unknown)P = k

P = P × k

P = P × k

.

.

.P = P × k (for the i’th statement)...P = P × k (for the N ’th statement)PRINT(P)END OF ALGORITHM

Here is the flow chart for the very, VERY, dumb way:


START

N FIXED NUMBERk FIXED NUMBERP IS UNKNOWN

.

.

.

.

.

.

P = k

P = P * k

P = P * k

P = P * k

P = P * k

STOP

PRINT P

N statements—lots of repetition of the same statement

A better (but still dumb) way:

START k is an integer or a real (fixed), N is an integer (fixed), P = 1 is a number thatwill contain the product (initialized to 1) and i is an integer (initialized to 0) used asan “index” or “counter” to indicate how many times we have passed through the loop.

Start of loop:


• i = i + 1 (“increment” the counter)

• P = P × k (“accumulate” the sum)

• IF i = N jump to End of loop:

• jump to Start of loop:

End of loop:PRINT(P)

END OF ALGORITHM

Here is the flow chart for the better way:

STOP

i = N ?

i = i + 1P = P * k

YES

PRINT P

NO

START

N FIXED NUMBERk FIXED NUMBER

i = 0; P = 1

Only a few statements to express the same idea.

A much, MUCH better way: Al-Kashi (1390 - 1450 AD)• al’Kashi, Ghiyath al’Din Jamshid Mas’ud• Born in Kashan (Iran), died in Samarkand (now called Uzbekistan)• “Self-proclaimed” inventor of decimal fractions• Calculated π to 16 decimal places• Theory of numbers and computation• “First” discovered the binomial theorem• Inventor of a calculating machine to predict lunar orbits• Discovered the following algorithm in 1414 AD:


START k is an integer or a real (fixed), N is an integer (fixed), P = 1 is a number thatwill contain the product (initialized to 1)

Start of loop:

IF (N is even)N = N/2

k = k × k

ELSE (i.e. N is odd)

N = N - 1

P = P × k

IF (N < 1) jump to End of loop:

ELSE (i.e. N ≥ 1) jump to Start of loop:

End of loop:

PRINT(P)

END OF ALGORITHM

This is the flow chart for the Al-Kashi Method:

N = N - 1P = P * k

YES

NON even

?

STOPPRINT P>-N 1 ?

YES

N = N / 2k = k * k

NO(N < 1)

START

N FIXEDk FIXED

P = 1 (PRODUCT)

Danger, danger: The algorithm redefines N and k.


3.9 An aside on computer architecture

3.9.1 What does S = S + 1 mean?

If you’re doing math, it means nonsense! It is the same as saying 0 = 1!

If you’re writing a computer program it means something else and describing what is meansrequires a small “aside” on computer architecture, the real “nuts and bolts” of what is goingon inside a computer.

First some definitions. See the figure below.

MEMORY The memory is the physical place in the computer where bit patterns are stored.Memory is best thought of as a stack of “words”, 32-bit bit patterns.

VARIABLE A symbol for a certain bit pattern (a 32-bit integer, say) that the programmercan define and change at will in a program. The “symbol” (S in this case) really refersto the location in memory where the bit pattern is stored.

CONSTANT A symbol for a certain bit pattern that the programmer can define and usebut not change.

CPU The Central Processing Unit in a computer. The CPU is the “boss”. It follows thesteps in a program and tells all the other parts of a computer what they should bedoing. It is the “controller”.

REGISTER A special place in the CPU where bit patterns are stored.

ADDER A functional unit in a computer (usually located right on the CPU chip) thatperforms additions on 32-bit bit patterns.

CPU

ADDER

R4R3R2R1

00000000000000000000000000000001

00000000000010010000100000100000

0000000000000000000000000000000000000000000000000000000000000000

0000000000000000000000000000000000000000000000000000000000000000

00000000000000000000000000000000

C:

S:

MEMORY

3.10. PROBLEMS 41

In a computer program, the statement: S = S + 1 gets “translated” into the followingoperations that the CPU performs or controls.

FETCH S,R1 Copy the bit pattern that is stored at the location referred to by S and putit in the 1st CPU register R1.

FETCH C,R2 Copy the bit pattern that is stored at the location referred to by C and putit in the 2nd CPU register R2. Previously in the program, the constant C would haveto have been defined and the bit pattern for C (00000000000000000000000000000001)would have to have been stored in C.

ADD R1,R2,R3 Copy the constants of the bit patterns stored in R1 and R2 into theADDER, do the binary arithmetic and copy the result into the 3rd CPU register R3.

STORE R3,S make a copy of the bit pattern in the 3rd CPU register R3 and place it inthe memory location referred to by S.

So, just the simple increment (add by 1) of a variable results in 4 operations performed bythe CPU. The preceding steps are exactly what an “assembly language” program would haveto do to interpret the statement S = S + 1. Remember that when we write statements likeS = S + 1 it is really a shorthand for a set of steps that a computer has to carry out. Wewill be clear in the future as to what is math and what is computer language. Math scriptwill be used for math (e.g. S = S+1) while type script will be used for computer instructions(e.g. S = S + 1).

3.10 Problems

1. The trinity of algorithmic design

What are the three most important capabilities that are needed to construct algo-rithms?

The ability to declare variables

The ability to assign values to variables

The ability to pass control from one operation to another in a prescribed way

A way of making a “True” or “False” decision

The existence of int’s and float’s

The ability to do mathematical operations (e.g. +, - ×)The ability to go back to a previous operation


A “compiler” that can convert human-readable text into a language (such asbinary) that a computer can understand

2. What is an algorithm?

In 35 words or less, answer the question, “What is an algorithm?”.


The three “pillars” of algorithm design are 1) sequencing, 2) branching, and 3) looping.

In 25 words or less for each, describe what they mean.That is:What is sequencing? (25 words or less)What is branching? (25 words or less)What is looping? (25 words or less)

4. Toss your cookies at this one!

In this Chapter, a recipe for Bielajew’s Sunday Morning Internationally-famous pan-cakes was given—in the form of pseudocode. Describe another recipe for a dish thatyou like—in the form of pseudocode. It has to have a Start, a Stop, more than 5 Steps,at least one Decision (or branch) point, and at least one Loop.

5. In the 70’s, man, if you were not hip, you were ...

Here is an algorithm in pseudo-code:

STARTOUTPUT “Enter a positive integer”INPUT Ni = 1WHILE (i× i < N)

i = i+ 1END WHILEIF (i× i = N)

OUTPUT “YES”ELSE

OUTPUT “NO”END IFSTOP

For which of the following inputs will the algorithm say “YES”?

(a) 28

(b) 36

3.10. PROBLEMS 43

(c) 54

(d) 81

What property of the input is the algorithm determining?

6. Summing the squares

Describe an algorithm in either pseudocode or flowchart form, to sum the squares ofthe integers up to and including some N , where N is an input to the algorithm. Thatis,

S = 12 + 22 + 32 + · · ·+N2

where S is the sum. You must express your algorithm in terms of a loop that is traversedN times. (There is a closed mathematical expression S = N(N + 1)(2N + 1)/6 thatyou must not exploit in your algorithm. Pretend the closed mathematical expressiondoes not exist.) Make sure to declare and initialize all the variables you use.

7. Don’t you love triangles?

A number N is called “triangular” if you can take N little asterisks and make anequilateral triangle from them. For example, here are the first few triangular numbers:

* * * * *

* * * * * * * * * *

3 * * * * * * * * * * * *

6 * * * * * * * * * * * *

10 * * * * * * * * * *

15 * * * * * *

21

Here is an algorithm that is attempting to detect triangularity in the input. But it hasa bug. Fix the algorithm.

STARTOUTPUT “Enter a positive integer”INPUT Ni = 1S = 0WHILE (S < N)

i = i+ 1S = S + i

END WHILEIF (S = N)

OUTPUT “Triangular”


ELSEOUTPUT “Not triangular”

END IFSTOP

8. De-Gauss this!

In this chapter we learned Mr. Gauss’s formula for the computation of the Nth trian-gular number:

1 + 2 + 3 + ...+N =N(N + 1)

2

Here is another algorithm attempting to detect triangularity of the input. Do youthink this one is OK? If not, fix it.

STARTOUTPUT “Enter a positive integer”INPUT Ni = 1WHILE [(i× (i+ 1))/2 < N ]

i = i+ 1END WHILEIF [(i× (i+ 1))/2 = N ]

OUTPUT “Not triangular”ELSE

OUTPUT “Triangular”END IFSTOP

9. The oscillating sign

Write an algorithm in pseudocode that accepts a positive integer N and outputs a 1(one) if N is even and -1 (minus one) if N is odd. Hint: -1 multiplied by itself an evennumber of times is +1, while -1 multiplied by itself an odd number of times is -1.

10. Summing an oscillating series

Describe an algorithm in both pseudocode and flowchart form, to perform the followingsum

S = 1− x+ x2 − x3 + x4 · · ·+ (−x)N

3.10. PROBLEMS 45

where x and N are inputs and S is the sum, an output. You must express youralgorithm in terms of a loop that is traversed N times. Express your algorithm interms of pseudocode and a flowchart.

11. Let’s talk about... the factorials of life

The factorial occurs in many places in mathematics and engineering applications andlife in general. The symbol for the factorial is the exclamation (!) mark. So, we wouldcall 3!, “three factorial”. Factorials are easy to calculate.1! = 12! = 2× 13! = 3× 2× 14! = 4× 3× 2× 1and so on.If N is a whole number,N ! = N × (N − 1)× (N − 2) · · ·2× 1.Note that N ! = N × (N − 1)!, a neat property of factorials.

You will describe an algorithm to calculate the factorial of N where N is any wholenumber greater than 1. You will describe an algorithm as if you were going to programit on a computer, that is, you will specify a set of instructions that even a machinecan understand. The algorithm should include instructions on how to start and end.You can assume that the machine on which the algorithm will be implemented canstore variables, loop, do branches, and do simple arithmetic. Express your algorithmin terms of pseudocode and a flowchart.

12. Decay and dissipation

You will describe an algorithm to calculate an approximation to the exponential func-tion, e−x. The following sum is a really good approximation to the exponential function,e−x, as long as the right hand side contains enough terms.

S = 1− x

1!+

x2

2!− x3

3!+

x4

4!+ · · ·

Let x and N , the number of terms on the right hand side, be inputs to the algorithmand S, the sum, be the output. Note the use of the factorials in the above sum. Ex-press your algorithm in terms of pseudocode and a flowchart. Note that the algorithmto raise x to a power is given in Chapter 3, Section 5 of “the book”.

Hint: The most efficient way to solve this problem is to rewrite S in the followingfashion:

S = s0 + s1 + s2 + s3 + s4 + · · ·


where s0 = 1 and note that

sn =(−x

n

)sn−1

13. What’s YOUR sine?

You will describe an algorithm to calculate the sine function, as if you were going toprogram it on a computer, that is, you will specify a set of instructions that even amachine can understand. The algorithm should include instructions on how to startand end. You can assume that the machine on which the algorithm will be implementedcan store variables, loop, do branches, and do simple arithmetic.

The following sum is a really good approximation to the sine function, as long as theright hand side contains more than 4 terms in the sequence shown:

S =x

1!− x3

3!+

x5

5!− x7

7!+ · · ·

Let x and the number of terms on the right hand side be inputs to the algorithm andS, the sum, be the output. Note the use of the factorials in the above sum. Expressyour algorithm in terms of pseudocode and a flowchart.

14. The rich get richer, the poor get poorer!

The financial investment firm of Getridge-Quigg, Ann Arbor, has a special deal forstudents. If you deposit your money with them they will give you 0.5% interest permonth (a large amount these days) but then they will charge you a $5.00 per monthservice charge applied against your account. If you have less than $5.00 in your account,they take the remainder and your account is closed. So, if you deposited $1000.00 andleave it there, your account will always have the same balance at the end of the month.If you deposit more than $1000.00 and leave it there you may gain. Deposit less than$1000.00 and your account may eventually be reduced to zero.

Write an algorithm that starts with an input of how much is to be deposited. Youintend to leave that money in the account and never add or withdraw from it. Youralgorithm will predict how many months it will take for you to double your moneyor to reduce your account to zero. If the balance in your account never changes, youhave to state that as well. Note that Getridge-Quigg first applies the interest and thenapplies the service charge. After every calculation, Getridge-Quigg rounds the resultto the nearest penny. For example, the monthly interest on a balance of $1001.00 is$5.01 while the monthly interest on a balance of $1000.99 is $5.00. You have to buildthis fact into your algorithm.

Express your algorithm in terms of pseudocode and a flowchart.

3.11. PROJECTS 47

3.11 Projects

1. Follow the bouncing ball

You will describe an algorithm as if you were going to program it on a computer, thatis, you will specify a set of instructions that even a machine can understand. Youralgorithm should specify its purpose, the input data (if any), the output data (if any),and the set of steps that comprise the algorithm itself. The algorithm should includeinstructions on how to start and end. You can assume that the machine on which thealgorithm will be implemented can store variables, loop, and do branches.

The problem is to determine the total distance a ball travels having been dropped froma height h (and traveling a total distance h on its way to the floor) and rebounding toa fraction of h which we will call ch (and traveling a total distance h+ ch). c is calledthe “coefficient of restitution”. The ball will then drop to the floor (total distance =h + 2ch) and bounce up to a height c2h (total distance = h + 2ch + c2h) and so on.You can assume for simplicity that the ball is a “point”, that is, its diameter is exactlyzero (a mathematical fiction to make our lives easier) and that c falls anywhere in theregion 0 < c < 1. Write an algorithm that determines the distance the ball travelsbefore coming to a rest. You can make any reasonable assumption you like about what“at rest” means for the ball. However, the assumption has to be “reasonable” andjustifiable either on the basis of numerics, mathematics or physics. This is a subtlepoint that is discussed below and is discussed in class.

This problem has an exact mathematical solution:

d = h1 + c

1− c

that you must not employ in your algorithm to determine the total distance traveled.You have to obtain the solution by developing an algorithm that sums individuallyevery bounce of the ball.

Some discussion:

A ball made from putty will have c = 0 and an ideal ball (which does not really existin reality) would have c = 1. A fresh “out-of-the-can” tennis ball has c = 0.5 or so, anold tennis ball has c = 0.25 or so. A ping-pong ball has c = 0.90 or so.

This problem is very much like Xeno’s paradox. The paradox is that mathematicallythe total distance traveled by the ball is finite (as long as c is less than 1) but math-ematically the total number of bounces is infinite! However, after many bounces therecovery distance becomes so small that accumulating them really does not matter inany practical sense any more. If digital computers could really represent real numbersto infinite precision then it would be possible to write an algorithm that accumulatesthe distance properly but it would never stop unless we introduced some artificial meansof stopping the execution. Part of the purpose of this exercise is to get you thinking


about this paradox, to realize that computers can only provide an approximation tomathematical fiction and physical reality.

Chapter 4

Getting started in C++

4.1 Simple input/output (I/O): A first program in C++

Here is a first program in C++. You can type it in yourself, or download it from the webfrom the lecture note distribution area for this class.

//File: goBlue.cpp

#include <iostream>

using namespace std;

int main(void)

{ cout << "Go Blue!\n";

// Multiple line statement

cout <<

"\n\n\t\"Go\t\t\\\n\t\t\t\tBlue!\"\a\a\a\r";

return(0);

}

There are several common features of every C++ program

Comments Comments start with /* and end with */. For example,

/* This is a comment */

49

50 CHAPTER 4. GETTING STARTED IN C++

Comments are put there for humans only. They do not affect or modify what thecomputer does. The computer does not care if you comment or not. But humansdo, especially your Lab Instructors and other people who hand out grades orpay your salary. Comments should explain what you are trying to accomplish andhow you are going about doing it. It is appropriate and usually correct to assumethat the person reading your code is a complete idiot and that you have to explainin detail what you are doing. After you have forgotten how or why you programmedyour own code, you will find your own comments very useful to understand what youhave done. The start of any program or program “segment” should begin with a fewlines of explanation, along with an identification of the programmer and the date itwas created. Put comments through the rest of the code wherever you think that thereader (aka the idiot) might need some explanation. Less is not more in this case. But,don’t go overboard too far. When in doubt, comment, comment, comment.

Comments can begin and end anywhere and can span more than one line.

// This is also a comment, from the "//" to the end of the line

which is useful when your comment is only one line or you want a comment to startsomewhere in a line and continue to the end of the line.

Use both kinds of comments and make your code readable.

Preprocessor directives Preprocessor directives instruct the compiler in advance of theactual compilation of your code. Preprocessor directives are indicated by a “#” be-ginning a line. In this case the preprocessor directive

#include <iostream>

instructs the preprocessor to include the contents of the ANSI standard library fileiostream. This header file contains other #include statements that call other libraryfiles that define the standards C++ input and output functions.

There are other common header files. We shall soon encounter cmath which is includedvia the preprocessor directive

#include <cmath>

which defines a lot of the common math functions. There are other types of prepro-cessor directives called “symbolic constants” and “macros” which we shall encounterlater.

If you want to have a look at these library files, look in the directory called

/usr/include/g++-3

where they reside. Boredom alert! This is not for the faint-of-heart!

4.1. SIMPLE INPUT/OUTPUT (I/O): A FIRST PROGRAM IN C++ 51

using namespace std; Tells the compiler that we will be using that standard “namespace”in the file that contains the line


All this really means is that standard library files and definitions are being used. It ispossible to define your own namespaces if you want to make sure that your variablesand definitions do not conflict with either the standard ones, or someone else’s, someoneworking on the same project. For this course, we shall be using the standard namespacefor everything we do. If you go on to work in a professional programming environment,writing software by teams, you will almost certainly be using your own namespaces.

int main(void){} All C++ programs begin processing at the first executable statement inmain. The “()”’s indicate that main is a block of code called a function. The int infront of main declares main to be a function that is expected to “return” an integer.(More on this later.) main is a special function that the operating system recognizesas the place to begin execution of a C++ program.

The lines of code in between the “{}”’s is called the function body. It is considered goodprogramming practice to indent the text in the function body by a certain amount ofspace. A common recommendation is 3 spaces which is reasonably pleasing to the eye.We will adopt this convention for this course. In our example, the function body startswith the statement

cout << "Go Blue!\n";

which instructs the C++ standard library function cout function to print the seriesof characters “Go Blue!” (not including the double quotes) to the screen and thenpositions the cursor at the beginning of the next line. This positioning is effected bya special sequence of characters “\n” called an escape sequence. Escape sequences arenecessitated when special characters are required or to “undo” the meaning of specialcharacters, like the double quote (”) in the cout statement above. Here is a list ofsome commonly used escape sequences employed in cout statements.

\n Newline. Move cursor to beginning of next line\t Horizontal Tab. Move cursor to next tab stop\r Return. Move cursor to beginning of current line\a Alert. Sound the bell\\ Print a backslash (\)\” Print a double quote (”)

As an extreme example contrived to show all the possibilities,


cout <<

"\n\n\t\"Go\t\t\\\n\t\t\t\tBlue!\"\a\a\a\r";

which throws 2 newlines, goes to the first tab stop, prints a double quote followedby “Go”, advances two more tab stops writes a backslash, throws a newline, goes tothe third tab stop and prints “Blue!” followed by a double quote, sounds the alarm3 times (You will probably only hear one beep, depending on your computer.) andissues a carriage return. Because a new line was not thrown, further keyboard inputcould overwrite the “Blue!” text. This just goes to show how far a professor will go tocontrive an example. Here is what the output looks like:

"Go \

red% Blue!"

Note that the statement starting with cout is terminated with a semicolon (;). Thisis a special character in C++ called a delimiter that signifies the end of a statement.There can be more than one statement per line in a C++ code or a single statementcan span several lines. Generally it is considered to be good programming practiceto put at most only one statement per line of code. If a statement is long, put it onseveral lines, space things nicely, and indent the carry-over lines 3 spaces relative tothe first line, like in the example above.

return(0); When this line is encountered, processing in the CPU returns to the operatingsystem. Since main is a function of the int type, an integer value must be “returned”to the process that called it, in this case the O/S. A return value of 0 usually meansnormal execution, by convention. A return other than 0 can be treated as an errorcondition by the O/S. For example, if your program had prompted for user input andthe user typed in something that the program was not prepared to process, one couldsignify this to the O/S by returning a 1, for example,

return(1);

You can have as many return’s as you need in a C++ code, just as you can have manystop points in an algorithm.

4.2 Compiling, linking, loading and running

Now that we have created a program, we have to tell the computer what to do with it! Afterall, computers only understand binary code and our program is written (coded) in terms ofcharacters.

To compile the program, which we will call goBlue.cpp, the following statement on a unixmachine attempts a compilation:

4.2. COMPILING, LINKING, LOADING AND RUNNING 53

red% g++ goBlue.cpp

The C++ compiler is another program that converts text instructions into binary code thata computer understands. Compilers have evolved over a long period of time. The originalcomputers (dating to the middle and late 40s) were actually programmed in binary code thatwas processed by the computer directly. This was an extremely laborious process. Early onit was recognized that it would be easier to program in a more human-compatible way, in atext-based format, and convert this into machine code using a compiler.

The origins of C++ date back to the late 60s and early 70s to the C language. The originalintent for C was to develop a language for writing computer operating systems. Operatingsystem (OS) code is special software for allowing users and programs to interact with datastored on a computer and instructs the operational units of a computer what to do withthis data. Later, in the 80s and 90s, C++ was developed as an evolution of the C language.(Any C++ compiler will compile C code.) The number of improvements is very great—toomany to mention.

C++ code is designed to be as hardware-independent as possible so that C++ code canmigrate between different computer architectures. C++ code is translated into assembleror assembly language which represents a text version (i.e. human-readable) version of bi-nary code. Assembly language is always machine dependent and addresses (speaks to) thecomputer hardware and its subcomponents. C++ is a higher-level language that deals withhardware and data in a more abstract way.

C and C++ have evolved beyond their original intent as OS-builders, although this role isstill very important. C and C++ are powerful general-purpose languages for most aspectsof computing. It is probably true that C and C++ are the most commonly-used computerlanguages in the world. C++ is certainly the one that is taught the most. There are otherhigh-level languages for special purpose applications in business and numerical computa-tion but probably C++ will supplant these as new business and numerical computationapplications evolve.

After an algorithm is developed through the conceptual, pseudocode and flowchart process, aC++ program (or other computer language program) is written. During a C++ compilationand execution, several steps occur:

Edit A human writes a computer program, e.g. goBlue.cpp Any editor will suffice. Com-mon unix editors are vi, emacs, pico, nedit or kwrite. Use whatever you like. Yourlab instructor will inform you of your options and probably ask you to stick with oneof these.

Preprocess, precompile Preprocessor directives, like #include and #define are handled.This can be inclusion of files or substitution of text throughout the C++ program.


Compile After the preprocessor, the compilation converts the code to object code alsoknown as binary code. There are actually several stages. The first is the parsing stagewhere the compiler checks the syntax of your code. If this stage is passed successfullythe next stage is the production of the assembly code from the C++ code which issyntax-error free. Then the assembly code is converted to binary code, also calledobject code.

Link After the object code is created the link stage is entered. At this point, referencesto functions that may have been referenced in the C++ program are established.These missing parts, in the form of object code, are either put in place directly wherereferenced (this is called static linking) or their transfer locations are established sothat processing can transfer to these locations and be resumed. This is called dynamiclinking. The file created by the linker is called the load module. By default, the loadmodule is called a.out by the unix operating system. A user can override this name ifdesired.

Load The loader takes the load module (usually located on disk) and puts it into memory ina place where the operating system expects to see executable code. On a unix systemthe loading phase is started when a user types the name of the load module residenton disk.

Run In the run phase, the binary code in the memory-resident load module is transferred tothe central processing unit (CPU) for execution. The load module contains instructionsfor the CPU that directs it to assign tasks to the various operational units of thecomputer. The load module also contains data and references to data that are tobe operated on. The run phase is where conceptual errors in a computer programare exposed. Code that has incorrect syntax, and syntax errors are quickly found bythe compiler’s parser, are usually easy to correct. Code that has correct syntax butincorrect design are much harder to find. This is where pseudocode and flow chartscan come to the rescue.

This model of editing, preprocessing, parsing, object-code generation, linking, loadingand running is followed by all computer code-development environments, from the mostsimple “Go Blue!” C++ program to the most complex, like the telephone managementsystem, international banking systems and the national security systems—irrespectiveof the language it is written in. The following figure includes not only the code compi-lation process, but also describes the whole process of code development. It is general,and mostly independent of the computer language being employed.

4.3. DECLARING AND INITIALIZING VARIABLES 55

PSEUDO CODEFLOWCHART

THINK

FAILURE

FAILURE

SUCCESS!

STEAL AS MUCHAS POSSIBLE

COMPILE

TYPE IT IN / EDIT

2) PARSING

3) CREATES OBJECT

1) PRECOMPILING

BAD DESIGN

REFINEMENT

LINK / CREATELOAD MODULE

RUN

4.3 Declaring and initializing variables

Consider the following C++ program that does mathematical operations with two integers:

//File: integerMath.cpp

#include <iostream>



int main(void)

{ int i1 = 5; // Define and initialize

int i2 = 2; // Define and initialize

int iResult; // Define only

iResult = i1 + i2; // sum

cout << "5 + 2 = " << iResult << "\n";

iResult = i1 - i2; // difference

cout << "5 - 2 = " << iResult << "\n";

iResult = i1 * i2; // product

cout << "5 * 2 = " << iResult << "\n";

iResult = i1 / i2; // division

cout << "5 / 2 = " << iResult << "\n";

iResult = i1 % i2; // remainder

cout << "5 mod 2 is " << iResult << "\n";

return(0);

}

There are some new features of this program.

The statement

int i1 = 5;

is a declaration statement. The name i1 is an identifier for the variable i1. This declarationstatement specifies that i1 is an int, a signed integer. It also provides an initialization fori1, giving it the value of 5.

An identifier is a series of characters you can type at a keyboard that can consist of letters,digits, and underscores ( ). It is a bad idea to start an identifier with an underscore, eventhough it is legal, because most identifiers in libraries accessed by C++ use this. An identifiercan not start with a digit, however. According to the draft ANSI C++ standard, an identifiermay be any length. ANSI [American National Standards Institute] C++ is a built uponANSI C, and in ANSI C only the first 31 characters are required to be recognized by Ccompilers conforming to the ANSI standard for C. Different C++ compiler writers determinehow many characters in an identifier are significant. I recommend NOT using more than

4.4. INTEGER MATH IN C++ 57

31, for safety. Beyond the limit of 31 or greater, your compiler may treat the charactersbeyond it as significant or insignificant (i.e. ignores them). For absolute safety, use 31characters or less. Identifiers in C++ are case sensitive so that i1 and I1 are recognizedas different variables. Unless your program is very short, you should name your identifiersin some recognizable way—it cuts down on the amount of commenting you have to do. Ialso like “humpback” notation, numberOfStudents, rather than number_of_students. Notethat humpback identifiers start with lowercase letters (usually) and new words within theidentifier start with an uppercase letter. I find underscores in the middle of identifiers ugly,wasteful and hard to type.

The statement

int i2 = 2;

is another declaration statement that identifies the name i2 and declares it to be a signedinteger initialized to the value 2.

The statement

int iResult;

is another declaration statement that identifies the name iResult and declares it to be asigned integer. In this case the variable is not initialized.

4.4 Integer math in C++

The statement

iResult = i1 + i2;

contains two mathematical operators, = and +. The operator = is called the assignmentoperator . Do not call it the “equal sign”! The assignment operator assigns the result ofthe operation(s) to the right of it to the value of the identifier to the left of it, iResult inthis example. = is called a binary operator because it operates on only two quantities, thevariable to the left which gets the assignment and the quantity to the right that providesthe assignment (after the other mathematical operations are completed).

The operator + is called the addition operator . It is also a binary operator and provides themathematical sum of the quantities to its left and right.

There are 4 other binary operators to consider now, the subtraction, -, the product , *, thedivision, /, and the modulus, %. Of these the division requires special mention with regards


to its operation with integers. The integer division returns the integer part of the result. Forexample, 5/2 = 2, not 2.5. The remainder of a division can be obtained from the modulusfunction %. Thus 5 % 2 = 1.

The various mathematical operations, when employed within the same statement, followcertain ordering rules. These are called the rules of precedence. The *, / and % operationstake precedence over + and -. *, / and % have the same precedence, and + and - have thesame precedence as well. For operators with the same precedence, the order of operation isleft to right. For example,

1 + 10 * 5 - 9 / 3 = 1 + 50 - 3 = 48

If a programmer desires a different interpretation of the above statement or wishes to makethe operation clearer, parentheses may be employed. Parentheses take precedence over *, /,%, + and -. Inner parentheses are evaluated first. For example,

(1 + (10 * 5) - (9 / 3)) = 1 + 50 - 3 = 48

is operationally equivalent to the above. However,

((1 + 10) * (5 - 9)) / 3 = (11 * (-4))/3 = -44/3 = -14

is quite different.

Summarizing, the order of precedence is

Order Symbol Operation Comments0 () parentheses inner ones evaluated first1 (unary) - negation (unary minus) R→L evaluation2 * / % product, division, modulus L→R evaluation3 + - addition, subtraction L→R evaluation

The statement

cout << "5 + 2 = " << iResult << "\n";

prints the characters “5 + 2 = ” to the screen then the value of iResult and then sendsthe newline escape sequence. There are more flexible ways of printing things to the screenand we shall encounter these shortly. Note how we can “cascade” the << operator for cout.An equivalent way of accomplishing the same thing would be:

cout << "5 + 2 = ";

cout << iResult;

cout << "\n";

4.5. FLOATING POINT MATH IN C++ 59

which is a little harder to read. Break up your output statement like this only for neatnessand readability.

4.5 Floating point math in C++

Consider the following C++ program that does mathematical operations with two floats:

//File: floatMath.cpp

#include <iostream>


int main(void)

{ cout << "Input the 1st floating point number: ";

float f1; // Declare the 1st float

cin >> f1; //and read it in

cout << "Input the 2nd floating point number: ";

float f2;

cin >> f2;

float fResult = f1 + f2; // sum

cout << f1 << " + " << f2 << " = " << fResult << "\n";

fResult = f1 - f2; // difference

cout << f1 << " - " << f2 << " = " << fResult << "\n";

fResult = f1 * f2; // product

cout << f1 << " * " << f2 << " = " << fResult << "\n";

if (0 == f2)

{ cout << "Division by zero not permitted!\n";

return(1);

}

else

{ fResult = f1 / f2; // division

cout << f1 << " / " << f2 << " = " << fResult << "\n";

return(0);

}

}


There are some new features in the above code.

float f1;

declares the existence of a floating-point number identified as f1. A “float” is a representationof numbers like 1.0, 2.5, 0.3333... in (usually) 32-bit representation. How a computer doesthis we will leave as a mystery for the time being. However, the representation of mostfloating point numbers is inexact, about 1 part in 107 for floats and about 2 parts in 1016 fora “double”, a 64-bit representation of real numbers. Float and doubles also have maximumand minimum values defined and this is architecture dependent, that is, on the computeryou are using. Common “dynamic ranges” for floats are between about 10−38 and 1038 andabout 10−308 and 10308 for doubles. We’ll have a lot more to say about floats as we getdeeper into the course. Floats and doubles are pretty much central to the issue of computingin engineering and we will discuss the concepts as we require them.

Note that in C++ we can declare variables anywhere. Many programmers like to declarethem just before using them for the first time.

The statement:

cin >> f1;

inputs the float f1 from the keyboard.

4.6 The if/else if/else construct

The if/else if/else construct is one of the most important programming features of C++and common to many computer languages. It has a single-choice form called if, a double-choice form called if/else and multiple-choice forms if/else if and if/else if/else.

The general form of the if/else if/else is:if (logical expression 1){

STATEMENT BODY 1

}else if (logical expression 2){

STATEMENT BODY 2

}else if (logical expression 3){

STATEMENT BODY 3

4.6. THE IF/ELSE IF/ELSE CONSTRUCT 61

}.

.

.

else if (logical expression N + 1){

STATEMENT BODY N + 1

}else

{STATEMENT BODY N + 2

}There are many forms of the if/else if/else construct:

• The else if (logical expression) { STATEMENT BODY } is optional

• The else { STATEMENT BODY } is also optional, independent of the above

• If the STATEMENT BODY consists of one and only one statement, then the surrounding{}’s are optional. (It can help with clarity of programming to keep them, however.Their use is highly recommended.)

• Note the use of indenting, optional but highly recommended

• At most one and only one of the STATEMENT BODY ’s can execute within an if/else

if/else construct can be executed

• Processing control passes to the statement immediately following the if/else if/else

construct after execution of one of the STATEMENT BODY ’s.

• If the if/else if or if form is used, it may happen that none of the logical condi-tions is satisfied. In this case processing control passes to the statement following theconstruct.

• if/else if/else constructs can be nested within each other

This is the flow chart for the if construct:


?

F

T

This is the flow chart for the if/else if construct:

?

?F

T

T

F

4.7. LOGICAL EXPRESSIONS 63

This is the flow chart for the if/else if/else construct:

?

?F

T

T

F

4.7 Logical expressions

So, what is a logical expression???

A logical expression is something that evaluates to TRUE or FALSE. FALSE has a numericvalue of the integer 0. TRUE has a numeric value of integer non-zero.

For example:

if (1)

cout << "This would always print,\n";

else if (0)

cout << "whereas this would NEVER print\n";

Here are some of the logical operators and their order of precedence


Order Symbol Operation Comments0 () parentheses inner ones evaluated first1 (unary) ! negation (unary NOT) R→L evaluation2 <=,< less than or equal, less than L→R evaluation2 >,>= greater than, greater than or equal L→R evaluation3 ==,!= equivalent to, not equivalent to L→R evaluation

Let’s return to the example of solving the quadratic equation, Ax2 + Bx + C = 0 that westudied in flowchart and pseudocode form in Lecture 3. If we look at the pseudocode onpage 7 of Lecture 3 we see that the if’s are nested and we would probably, at first try, writea code that looked something like the following, using only if’s and else’s:

//File: quadratic.cpp

#include <iostream>

#include <cmath>


int main(void)

{ float a,b,c; // Declare the constants...

cout << "\n"

"We will solve for x in Ax*x + B*x + C = 0\n"

"=========================================\n"

"\n";

cout << "Input A,B,C: ";

cin >> a >> b >> c; //...and read them in

cout <<

"Calculating roots (x) of the equation\n"

<< a << "*x*x + " < b*b - 4*a*c) //b*b - 4*a*c is less than 0

{ cout << "No solution since b*b-4*a*c < 0\n";

return(0);

}

else // b*b - 4*a*c is >= 0

{ if (0 != a)

{

cout <<

"Sol’n 1 = " << (-b + sqrt(b*b - 4*a*c))/(2*a) << ", "


"Sol’n 2 = " << (-b - sqrt(b*b - 4*a*c))/(2*a) << "\n";

return(0);

}

else // b*b-4*a*c is >= 0 and a = 0

{ if (0 != b) // b*b - 4*a*c is >= 0, a = 0, b != 0

{ cout << "One sol’n since a = 0. Sol’n = " << -c/b << "\n";

return(0);

}

else // b*b - 4*a*c is >= 0, a = 0, b = 0

{ cout << "No sol’ns since a and b = 0.\n";

return(0);

}

}

}

}

However, by using else if’s we can make the code look less “nested” and complicated. Thenesting is really still there and we have to remember that when we pass into an else if

that the if condition above it is still false! However, else if’s make the coding look neaterand easier to read.

//File: quadratic2.cpp

#include <iostream>

#include <cmath>


int main(void)

{ float a,b,c; // Declare the constants...

cout << "\n"

"We will solve for x in Ax*x + B*x + C = 0\n"

"=========================================\n"

"\n";

cout << "Input A,B,C: ";

cin >> a >> b >> c; //...and read them in

cout <<

"Calculating roots (x) of the equation\n"

<< a << "*x*x + " < b*b - 4*a*c) //b*b - 4*a*c is less than 0

{ cout << "No solution since b*b-4*a*c < 0\n";

return(0);

}

else if (0 != a) // b*b - 4*a*c is >= 0, a != 0

{ cout <<

"Sol’n 1 = " << (-b + sqrt(b*b - 4*a*c))/(2*a) << ", "

"Sol’n 2 = " << (-b - sqrt(b*b - 4*a*c))/(2*a) << "\n";

return(0);

}

else if (0 != b) // b*b - 4*a*c is >= 0, a = 0, b != 0

{ cout << "One sol’n since a = 0. Sol’n = " << -c/b << "\n";

return(0);

}

else // b*b - 4*a*c is >= 0, a = 0, b = 0

{ cout << "No sol’ns since a and b = 0.\n";

return(0);

}

}

4.7.1 Logical expressions with AND or OR

Here are more of the logical operators and their order of precedence:

Precedence Symbol Operation Comments0 (first) (,) parentheses, paired L→R evaluation

1 (unary) ! negation (unary NOT) R→L evaluation2 <=,< less than or equal, less than L→R evaluation2 >,>= greater than, greater than or equal L→R evaluation3 ==,!= equivalent to, not equivalent to L→R evaluation4 && logical AND L→R evaluation

5 (last) || logical OR L→R evaluation

The AND and OR operators work as follows. Imagine that two logical expressions evaluatesto either T (TRUE) or F (FALSE). This is the result of the && and ||’s operating on them.


(T && T) == T(T && F) == F(F && T) == F(F && F) == F(T || T) == T(T || F) == T(F || T) == T(F || F) == F

Sometimes these tables are difficult to remember, at least as abstract concepts. This ishelped somewhat if we substitute numerical values for T (1) and F (0).

(1 && 1) == 1(1 && 0) == 0(0 && 1) == 0(0 && 0) == 0(1 || 1) == 1(1 || 0) == 1(0 || 1) == 1(0 || 0) == 0

Once we do this we realize that the result of (A && B) is the lower value of either A or B. Theresult of (A || B) is the higher value of either A or B. It is no more complex than that! Andthis is, in fact, how the electronic circuitry works in a computer during the interpretation oflogical expressions.

Some examples:

int i = 5;

if (0 i)

cout < i || 10 <= i)

cout << i << " is either negative, or has more than one digit\n";

4.7.2 Mixed arithmetic and logical expressions

It is possible to mix arithmetic and logical expressions. Usually it is done where a logicalexpression is expected, such as within an if/if else/else construct. It is also possible tomix these operators elsewhere, where a regular statement is expected. Don’t do it—unlessyou really, really have to for some purpose!


When logical operators and arithmetic operators are mixed in a single expression, the orderof precedence is as follows. Above all, the contents of parentheses are evaluated starting withthe innermost. Then the arithmetic operators get interpreted in the order already discussed.Finally, the logical operators get interpreted in the order already discussed. Here is the tablefor precedence for both arithmetic and logical expressions.

Precedence Symbol Operation Comments0 (first) ( ) parentheses, paired L→R evaluation, innermost first

1 (unary) - ! negation R→L evaluation2 * / % multiply, divide, modulus/remainder L→R evaluation3 + - add, subtract L→R evaluation4 <=,< less than or equal, less than L→R evaluation4 >,>= greater than, greater than or equal L→R evaluation4 ==,!= equivalent to, not equivalent to L→R evaluation5 && logical AND L→R evaluation6 || logical OR L→R evaluation

7 (last) = assignment R→L evaluation

When in doubt, use parentheses. When not in doubt, use parentheses anyway. Theperson reading your code may have forgotten all but the first rule of precedence, parenthesesare interpreted first. This will make your code more readable and understandable.

In fact, that is the convention I have adopted in almost all of my programming, put paren-theses around every logical sub-expression in any logical expression that contains more thanone logical operators. I also put parentheses around any arithmetical sub-expression in amixed arithmetic/logical expression.

Here’s an example of a mixed arithmetic and logical expression. Use the rules of precedencestated above to see how the following statement is evaluated:

if (i*j == i*i + j*j || i - j <= i*i - j*j && i + j <= i*i + j*j) ...;

The ordering of && and || is also quite important. For example, using the rules of precedencethe above could collapse to 1 || 1 && 0 which is 1 or TRUE. However, if you mistakenlythought that the || is evaluated first, you would guess that the answer should be 0 or FALSE,and that is the wrong answer in this case.

4.8. PROBLEMS 69

4.8 Problems

1. An essay-type question? In a computer1 class!

Answer the following questions in your own words...

(a) What is an algorithm?

(b) Not counting the ability to START and STOP, what are the 3 functional build-ing blocks that are used to construct any algorithm.

(c) Give one example each of these 3 functional building blocks

(d) Name 3 ways in which algorithms may be described.

(e) Give one example each of the above 3 ways in which algorithms may be described.

2. Basic Arithmetic

Write down the value that is assigned to the variable on the left-hand side of each ofthese expressions. The variables i and j are int’s; x and y are float’s.

i = 1 + 2*3/4+5;

x = 1.0 + 2.0 + 3.0/4.0 + 5.0;

j = (1 + 2)*3/4 + 5;

y = (1.0 + 2.0)*3.0/4.0 + 5;

y = 1 + 2 + 3/4 + 5;

x = 1 - (static_cast<int>(6*5/4.0*3))/2;

3. Basic Logical Operations

Do the following logical expressions evaluate to TRUE (1) or FALSE (0)?i is an int that can take on any value.

(1 || 1 && 0 || 0)

(0 > i && 10 < i)

(0 > i || -10 < i)

4. Mixed Logical and Mathematical Operations

Do the following mixed mathematical-logical expressions evaluate to TRUE (1) orFALSE (0)?i is an int that can take on any value. (Hint: No matter what i is, 2*i + 1 is an oddnumber and 2*i is an even number.)

1Hey! Unless you’ve been on the moon for the last two weeks...This is a course about thinking andmaking things work! Writing computer program is just one way (and a good way for Engineers!) to teachthis.


(2*i + 1)

(4*i%2 && 1)

(2*i + 1 || i/2)

5. Mind Your P’s, Q’s and Semi-Colons

Which of the following are valid statements?

int i, j, k = 0;

int i, double j;

int i0 = 4, f = -1, k=0;

int i; float _0j;

int i j k l m;

int The, Wolverines, Are, Rated, Number, 3;

float f, F, _000000, Wolverine;

int i,j ; i == 1 && j || 3%2;

int static_cast<float>(N);

i = -i + -j;

i + j = f;

x = x*x + 3;

y = x^2 + 3;

y = a x + b;

z = x + y123;

z = x + 123y;

for(i = 0,i <= 10,i = i + 1) cout << "OK?\n";

while(thisIsTrue) cout << "OK?\n";

if (((x < 0) && (y > 3) || (z == 2)) cout << "OK?\n";

if (a != b && c > d && e < f && g == h) cout << "OK?\n";

6. Predict the Output

The examples of code given below all compile correctly. Predict the output that thecode would produce if you compiled and ran it.

• int N = 0;

if (N = 0)

{ cout << "The conditional expression was TRUE\n";

}

else

{ cout << "The conditional expression was FALSE\n";

}

• int i = 0;

if (0 != i)

4.8. PROBLEMS 71

cout << "Statement 1\n";



• int j = 0;

if (0 < j);

{ j = j + 1;

}

cout << "j = " << j << \n";

7. This question is iffy, REALLY iffy

Dreaming up your own examples, write syntactically correct C++ code for the follow-ing. I have provided the answer to the first question to give you an idea of what I amlooking for. You do not have to declare or initialize any of the variables you use.

(a) An example of the use of the if. If the condition is TRUE then one and onlyone statement is executed.

A correct answer is: if (i < N) f = f*i;

(b) An example of the use of the if. If the condition is TRUE then exactly twostatements are executed.

(c) An example of the use of the if and the else if. If the if condition is TRUEthen exactly two statements are executed. If the else if condition is TRUEthen one and only one statement is executed.

(d) An example of the use of the if and two else if’s.

(e) An example of the use of the if, two else if’s and the else.

8. Location, location, location

Write a program that prompts its user for two int’s, the first representing an x-coordinate and the second representing a y-coordinate. Then the program prints outwhere the (x, y) point is located, based on the following scheme.

x > 0 y > 0 upper right quadrant (URQ)x > 0 y < 0 lower right quadrant (LRQ)x < 0 y > 0 upper left quadrant (ULQ)x < 0 y < 0 lower left quadrant (LLQ)x = 0 y > 0 upper half (UH)x = 0 y < 0 lower half (LH)x > 0 y = 0 right half (RH)x < 0 y = 0 left half (LH)x = 0 y = 0 in the center (C)


Feel free to use abbreviations, like URQ, for upper right quadrant.

Here’s an example of how it works:

% a.out

Enter two int’s, representing x and y: 1 1

Upper right quadrant (URQ)

%

4.9. PROJECTS 73

4.9 Projects

1. Your first programming problem

You have to create an entire code that will compile correctly, accept inputs and provideoutputs and stops. As for coding style, follow Appendix A.

Your program will do the following:

(a) Ask the user to input two non-negative integers.

(b) If the user inputs a negative integer, the program issues an error message andstops.

(c) If the two numbers are equal, output is provided to inform the user. Otherwisethe user is informed which number is larger and which is smaller.

(d) The program outputs the ratio of the larger number to the smaller number unlessthe smaller number is zero. If the smaller number is zero, a message to this effectis output.

(e) The program outputs the quadrature sum of the two numbers, that is,firstInteger * firstInteger + secondInteger * secondInteger.

(f) If the smaller number is zero, the program stops. Otherwise, the program checksif the smaller number is a perfect divisor of the larger number. This means thereis no remainder after division. For example, 2 is a perfect divisor of 8, but not 7.Output is provided to inform the user whether or not a perfect divisor was found.

(g) Program stops.

The above set of instructions is really a coarse pseudocode for your program. Read theinstructions very carefully.

Here are some example uses of the program. Your program should work exactly thesame.

> a.outInput the 1st integer: -1Input must be non-negative. Try again!

> a.outInput the 1st integer: 1Input the 2nd integer: -1Input must be non-negative. Try again!

> a.outInput the 1st integer: 1Input the 2nd integer: 1


The two numbers are the same!The ratio is 1, the quadrature sum is 21 is a perfect divisor of 1

> a.outInput the 1st integer: 10Input the 2nd integer: 0The bigger number is 10, the smaller number is 0The smaller number is 0, so no ratio is calculated, the quadrature sum is 100

> a.outInput the 1st integer: 0Input the 2nd integer: 10The bigger number is 10, the smaller number is 0The smaller number is 0, so no ratio is calculated, the quadrature sum is 100

> a.outInput the 1st integer: 7Input the 2nd integer: 4The bigger number is 7, the smaller number is 4The ratio is 1, the quadrature sum is 654 is not a perfect divisor of 7

> a.outInput the 1st integer: 4Input the 2nd integer: 7The bigger number is 7, the smaller number is 4The ratio is 1, the quadrature sum is 654 is not a perfect divisor of 7

> a.outInput the 1st integer: 8Input the 2nd integer: 2The bigger number is 8, the smaller number is 2The ratio is 4, the quadrature sum is 682 is a perfect divisor of 8

> a.outInput the 1st integer: 2Input the 2nd integer: 8The bigger number is 8, the smaller number is 2The ratio is 4, the quadrature sum is 682 is a perfect divisor of 8

4.9. PROJECTS 75

2. Leap year calculation

The rules to calculate if it is a leap year are:

(a) A year divisible by 4 is a leap year (2004, 2008...), unless

(b) it is also divisible by 100 (2100, 2200...) in which case it is not a leap year.

(c) There is an exception. A year divisible by 400 is a leap year (2000, 2400...).

Write a program that will accept a positive int, the year to be tested, from a user.If the user inputs a 0 or negative number, the program quits without doing anything.Otherwise, it will print a message saying whether or not the year is a leap year. Hereis an example of how the program works.

unix-prompt> a.out

Input a number for the year (>= 1): -1

Sorry, your year must be greater than 1.

unix-prompt> a.out

Input a number for the year (>= 1): 2001

The year 2001 is not a leap year

unix-prompt> a.out


The year 2004 is a leap year

unix-prompt> a.out


The year 2100 is not a leap year

unix-prompt> a.out


The year 2400 is a leap year

3. When in Rome...

The following table shows how Roman numerals are formed:

Arabic Roman Arabic Roman Arabic Roman Arabic Roman1 I 10 X 100 C 1000 M2 II 20 XX 200 CC 2000 MM3 III 30 XXX 300 CCC 3000 MMM4 IV 40 XL 400 CD5 V 50 L 500 D6 VI 60 LX 600 DC7 VII 70 LXX 700 DCC8 VIII 80 LXXX 800 DCCC9 IX 90 XC 900 CM


A Roman number is written in decades starting with the largest decade on the leftand then going to the right. For example, the number 1999 would be represented asMCMXCIX. The largest number that can be represented by Roman numerals is 3999,or MMMCMXCIX. Write a program that will accept a positive int, between 1 and3999 inclusive. If the user inputs something outside this range, the program quitswithout doing anything. Otherwise, it will print the conversion to Roman numerals.Here is an example of how the program works.

unix-prompt> a.out

Input an integer between 1 and 3999 inclusive: 0

Sorry, your integer must be between 1 and 3999 inclusive.

unix-prompt> a.out


Sorry, your integer must be between 1 and 3999 inclusive.

unix-prompt> a.out


MCCCXXXVIII

unix-prompt> a.out


MMMDCCCLXXXVIII

unix-prompt> a.out


LII

4. This is a 5-bit calculation

Write a program that will accept 5 bits (0 or 1) in the following order: b0, b1, b2, b3, b4representing a 5-bit bit pattern b4b3b2b1b0. (Note the ordering.) Write a program that:

(a) Determines if the 5-bit bit pattern b4b3b2b1b0 represents an odd or even unsignedinteger.

(b) Determines if the bit pattern represents a negative or positive signed integer in5-bit two’s complement representation.

(c) Prints the unsigned integer representation.

(d) Prints the bit pattern’s bit complement b̃4b̃3b̃2b̃1b̃0 (1’s become 0’s and 0’s become1’s).

(e) Prints the signed integer representation assuming 5-bit two’s complement repre-sentation.

Here is an example of how the program works.

4.9. PROJECTS 77

unix-prompt> a.out

Bit b0: 0

Bit b1: 1

Bit b2: 0

Bit b3: 1

Bit b4: 0

5-bit bit pattern is: 0 1 0 1 0

Bit pattern represent an even unsigned integer

Bit pattern represent a positive signed integer

Unsigned integer representation is: 10

Complement 5-bit bit pattern is: 1 0 1 0 1

Signed integer representation is: 10

unix-prompt> a.out

Bit b0: 1

Bit b1: 1

Bit b2: 1

Bit b3: 1

Bit b4: 0


Bit pattern represent an odd unsigned integer

Bit pattern represent a positive signed integer



Signed integer representation is: 15

unix-prompt> a.out

Bit b0: 1

Bit b1: 1

Bit b2: 1

Bit b3: 0

Bit b4: 1



Bit pattern represent a negative signed integer



Signed integer representation is: -9

unix-prompt> a.out

Bit b0: 1

Bit b1: 0


Bit b2: 0

Bit b3: 0

Bit b4: 1



Bit pattern represent a negative signed integer



Signed integer representation is: -15

5. Is binary arithmetic iffy

We return to binary arithmetic of 5-bit bit patterns that was considered in one ofthe problems in the previous chapter. We imagine a machine that can only representintegers in 5-bit bit strings. However, this time you will write a program that emulates5-bit binary arithmetic using actual int’s in C. Remember that there is no 6’th orhigher bit. The result of any calculation that would carry over into the 6’th bit isthrown away.

The program you will write will do two things, depending on the choice of the personusing your program:

(a) multiply a 5-bit bit pattern by a 2-bit bit pattern, for example, 10101× 11, or

(b) add two 5-bit bit patterns, for example, 10101 + 10011.

The user of your program will provide the bit patterns that are to be added or multi-plied. Your code will output the 5-bit bit pattern that is the result of the multiplicationor addition.

Examples of use:

Here is an example of running the code:

> a.out

What do you want to do?

Type a 0 if you want to add two 5-bit strings

Type a 1 if you want to multiply a 5-bit string by a 2-bit string

(0 or 1): 1

Enter the 5-bit binary number to be multiplied: 1 0 1 0 1

Enter the 2-bit multiplier: 1 1

4.9. PROJECTS 79

10101

x 11

------

...is the same as the addition...

10101

+01010

------

11111

>

Here is another example of running the code:

> a.out

What do you want to do?

Type a 0 if you want to add two 5-bit strings

Type a 1 if you want to multiply a 5-bit string by a 2-bit string

(0 or 1): 0

Enter the 1st 5-bit binary number to be added: 1 0 1 0 1

Enter the 2nd 5-bit binary number to be added: 1 0 0 1 1

10101

+10011

------

01000

>

6. Change for the good

You will write a program that will accept an input, a real number between 0.00 and19.99 inclusive. This represents the “change” in dollars from the purchase of an itemusing a $20 bill. You will write a C++ program that will calculate the most efficientform that the change may be given in terms of $10, $5, and $1 bills plus quarters,dimes, nickels and pennies.

Here’s an example of how it should work. Note that the program was run twice. Thefirst time the input was 19.94, the second time 16.26. Note how the program correctlyuses singular (One penny) and plural (Four $1 bills).


unix-prompt> a.out

Input a dollar amount between 0.00 and 19.99 inclusive: 19.94

Change of $19.94 is best given as:

One $10 bill

One $5 bill

Four $1 bills

Three quarters

One dime

One nickel

Four pennies

unix-prompt> a.out

Input a dollar amount between 0.00 and 19.99 inclusive: 16.26

Change of $16.26 is best given as:

One $10 bill

One $5 bill

One $1 bill

One quarter

One penny

unix-prompt>

7. Base 10 to binary conversion

Write a C++ program that will accept any integer between 0 and 127, inclusive, andconvert it to binary. You may not use loops of any kind and your solution must NOTresemble the following:

if (decimal == 0) cout << "0";

else if (decimal == 1) cout << "1";



.

.

.



Here’s an example of how it should work for an input of 126:

unix-prompt> a.out

Input a positive int between 0 and 127 inclusive: 126

4.9. PROJECTS 81

Decimal 126 converted to base 2 is 1111110

unix-prompt>

Hint: Depending on your algorithm, you may find it useful to use the escape sequence“\b” which causes the cursor to move backwards one space.

8. Something loopy to think about

This is a discussion question, not a programming assignment, because an elegant so-lution to this requires the syntax introduced in the next chapter. It is meant to makeyou think about loops, the topic of the next chapter. How would you write a C++program that could convert any unsigned integer to any base between 2 and 9? Usingloops, this actually easier to do than the solution to the previous problem!


unix-prompt> a.out

Input the base to convert to between 2 <= b <= 9: 7

Input any positive unsigned int: 126


unix-prompt>


Chapter 5

Loops

Loops are the third and final programming element we need to write any algorithm. TheC++ language provides three “easy” ways to do this, the while, do/while and for loops.

5.1 The while loop

The general form of the while loop is:while (logical expression){

STATEMENT BODY 1

}Some features of the while loop:


• Note the use of indenting.• Processing control passes to the statement immediately following the while constructafter the condition tested by the logical expression fails.

• while loops (in fact, any repetition loop) can be nested within each other• The logical expression can mix arithmetic and logical operators to make for compactcoding. (This is not really recommended for novice programmers and often makes codehard to read.)

• If the logical expression is false (even the first time), the STATEMENT BODY within thewhile loop is never executed.

• If the condition which is tested is TRUE and neither the STATEMENT BODY nor thelogical expression modifies it so that it would become false, then it will run forever.This is called a “forever” loop, while(1){}. Forever loops are bad, anti-social things.

83

84 CHAPTER 5. LOOPS

It puts your computer in a state known formally as “hung”, as in, “Jeepers, I thinkmy computer is hung.” It is NOT a compliment! Hung computers are put out of theirmisery by a variety of desperate measures, increasing in severity. At the very worst,the power has to be shut down, a really nasty, last resort for a hung computer.

This is the flow chart for the while loop:

?

T

F

STATEMENT BODY

Here’s an example that tests Gauss’s summation formula:

//File: gauss.cpp

#include <iostream>


int main(void)

{ cout << "Input N: ";

int N;

cin >> N;

int i = 0, Sum = 0;

while (i < N)

{ i = i + 1;

5.1. THE WHILE LOOP 85

Sum = Sum + i;

}

cout <<

"Sum of " << N << " digits is " << Sum

<< "...according to Gauss: " << N*(N+1)/2 << "\n";

return(0);

}

Here’s a more compact (but dangerous way) to do it:

//File: gaussAgain.cpp

#include <iostream>


int main(void)

{ cout << "Input N: ";

int N;

cin >> N;

int i = 0, Sum = 0;

while (i = i + 1 <= N) Sum = Sum + i; //NO, NO, NO!

cout <<

"Sum of " << N << " digits is " << Sum

<< "...according to Gauss: " << N*(N+1)/2 << "\n";

return(0);

}

This is dangerous because if you forgot the parentheses around the i = i + 1 part of theconditional, you would generate an infinite loop! Don’t believe me? Try it! Um, wait, thisis a forever loop! OK, try it, but <CNTL>-C out of it after a few seconds.

The point is—try to make your coding as easy to interpret as possible. Don’t be too clever.Don’t save space just for the purpose of saving space at the expense of clarity. In the old daysof computing, memory and disk space were very expensive and precious. Programmers pridedthemselves at making things very compact. Those days are, happily, long gone. However,there still are people who do this. Maybe they think that an electron is as expensive as aHiggs’ boson1.

1Electrons are essentially free. There are countless numbers of them in everyday things, like water. The

86 CHAPTER 5. LOOPS

5.2 The do/while loop

The general form of the do/while loop is:do

{STATEMENT BODY 1

}while (logical expression);

Some features of the do/while loop:

• Do not forget the “;” after the condition!• If the STATEMENT BODY consists of one and only one statement, then the surrounding{}’s are optional. (It can help with clarity of programming to keep them, however.Their use is highly recommended.)

• Note the use of indenting.• Processing control passes to the statement immediately following the do/while con-struct after condition tested by the logical expression fails.

• do/while loops (in fact, any repetition loop) can be nested within each other.• The logical expression can mix arithmetic and logical operators to make for compactcoding. (This is not really recommended for novice programmers and often makes codehard to read.)

• The STATEMENT BODY is executed at least once even if the logical expression is falseinitially.

• If the condition which is tested is true and neither the STATEMENT BODY nor the logicalexpression modifies it so that it would become false, then it will run forever. This iscalled a “forever” loop, do{}while(1);.

This is the flow chart for the do/while loop:

Higgs’ boson is very expensive. Physicists are spending billions to find it. It has not been found yet, butalmost! The first person to find it will win the Nobel Prize for physics.

5.2. THE DO/WHILE LOOP 87

?

F

T

STATEMENT BODY

The only thing that really distinguishes a do/while loop from a while loop is that thedo/while loop evaluates the conditional at the end of the STATEMENT BODY and the while

loop evaluates it at the beginning. while loops are often used when incrementing a counterand using (and usually updating) the counter within the statement body. do/while loopsare useful if you always want to process the STATEMENT BODY at least once. Because of this,loops that accept input from the keyboard (or a file) make good use of the do/while loopespecially when the input is terminated using a “sentinel”.

Here is an example of sentinel controlled output for calculating the average of student grades.The sentinel for stopping is when the input for a grade is less than zero. Even the cruelestprofessor does not hand out a negative grades (even though it is legal)! So, this is a goodsentinel to use for this application.

//File: grades.cpp

#include <iostream>


88 CHAPTER 5. LOOPS

int main(void)

{ int nStudents = 0, grade, sumGrades = 0;

do

{ cout << "Input a grade: ";

cin >> grade;

if (0 > grade) cout << "Grade < 0, end of input.\n";

else

{ nStudents = nStudents + 1;

sumGrades = sumGrades + grade;

}

}while (0 <= grade);

if (0 < nStudents)

cout << nStudents

<< " grades were read in. "

<< "The class average is: "

<< sumGrades/nStudents

<< "\n";

return(0);

}

Note that the above calculates the average as sumGrades/nStudents, the result of an integerdivision. This may not be completely fair as an average of, say, 79.9 should not really bereported as 79.

The way to “fix” this is to make the following changes:

1. Add the following #include in the pre-processor area of the code:

#include <iomanip>

2. Modify the cout statement as follows:

cout << nStudents

<< " grades were read in. "

<< "The class average is: "

<< setw(5) //Set field width to 5

<< setprecision(3) //3 digits of precision

<< static_cast<float>(sumGrades)/nStudents

<< "\n";

5.3. THE FOR LOOP 89

The code in its entirety can be found in the lectures area on the web. It is called grades1.cpp.

There are two new features in the cout statement. One is the introduction to the unary“cast” operator static_cast<float>(expression) which converts sumGrades temporarily(and internally) to a float and the resulting calculation takes place as if it were a divide usingfloats alone. It is not necessary to make the same conversion for nStudents. The divisionof a float by an integer or an integer by a float always results in a floating point calculationas the integer participating is “promoted” to a float.

The general form of the cast operator is

static_cast<type_name>(expression) //Don’t forget the parentheses!

to change expression to the type type_name. So, float’s can be changed to int’s and vice-versa. It is possible to change among other types of variables that we will encounter later.

The other new feature is the use of a “field width” and “precision” to modify the output ofthe float. The function setw(m) sets aside m spaces to print. The function setprecision(n)

prints n significant digits. These functions are defined in the iomanip (I/O stream manipula-tor) library file. Hence, we had to include this file, else the compiler would have complained.Note how these functions are employed. setprecision() only has to be set once. Thestream handlers remember it. However, setw() has to be set for every item!

There is another commonly used stream manipulator called fixed. We will not use it much.

5.3 The for loop

The for loop is similar to the while loop except that the syntax makes explicit the ini-tialization, the looping condition and the loop counter modification. This form is generallypreferred if the loop is to be performed in a systematic fashion, to a well-defined terminationcriterion with a regularly-spaced or easy-to-calculate loop counter indexing.

The general form of a for loop is:for (expression 1; logical expression; expression 2)

{STATEMENT BODY

}

Some features of the for loop:


90 CHAPTER 5. LOOPS

• Note the use of indenting.

• Processing control passes to the statement immediately following the for constructafter condition tested by the logical expression fails.

• for loops (in fact, any repetition loop) can be nested within each other.

• The logical expression is intended to provide the condition for which the executionof the for loop terminates. This logical expression can mix arithmetic and logicaloperators to make for compact coding. (This is not really recommended for noviceprogrammers and often makes code hard to read.)

• The STATEMENT BODY will not be executed if the logical expression is false initially.

• expression 1 is intended to initialize the loop counter (or index). It’s use is optional.However, in this case the initialization of any counters (should they be used) shouldbe done beforehand.

• The use of the logical expression is also optional. If not provided, the STATEMENTBODY is always entered. Another way of exiting the loop has to be provided otherwisethe loop is infinite.

• expression 2 is intended to increment the counter. The statement (or statements)are executed within the loop after the STATEMENT BODY. It’s use is also optional. If notspecified, the increments (or decrements) of any counters should be done within theSTATEMENT BODY.

• The two “;”’s in the definition of the for loop are NOT optional. Leaving one or bothout will result in a compilation error.

This is the flow chart for the for loop:


?

T

STATEMENT BODY

EXPRESSION 2

F

EXPRESSION 1

Here is an example that executes a count down and a “blast off”.

//File: blastOff.cpp

#include <iostream>


int main(void)

{ cout << "Estimate number of counter loops/second: ";

int loopsPerSecond;

cin >> loopsPerSecond;

cout << "How many characters wide is your screen? ";

int screenWidth;

cin >> screenWidth;

cout << "How many characters high is your screen? ";

92 CHAPTER 5. LOOPS

int screenHeight;

cin >> screenHeight;

for (int i = 10; i >= 0 ; i = i - 1)

{ // Next line should waste about 1 second

for (int j = 0; j <= loopsPerSecond; j = j + 1) ;

cout << i << "\n";

}

cout << "\a.\n.\n.\n.\nWe Have ignition!\a\n.\n.\n.\n";

//"Draw" a rocket

cout << " /\\\n"

<< " ||\n"

<< " ||\n"

<< " ||\n"

<< " ||\n"

<< "/||\\\n";

//Start of the vapor trail

cout << " **\n";

for (int j = 0; j <= loopsPerSecond/2; j = j + 1) ;

cout << "*****\n";

for (int j = 0; j <= loopsPerSecond/5; j = j + 1) ;

//Vapor trail expands

for (int i = 6; i <= screenWidth; i = i + 1)

{ for (int j = 1; j <= i; j = j + 1) cout << "*";

for (int j = 0; j <= loopsPerSecond/i; j = j + 1) ;

cout << "\n";

}

//Vapor trail at maximum

for (int i = 1; i <= screenHeight; i = i + 1)

{ for (int j = 1; j <= screenWidth; j = j + 1) cout << "*";

for (int j = 0; j <= loopsPerSecond/screenWidth; j = j + 1) ;

cout << "\n";

}

//Vapor trail contracts

for (int i = screenWidth; i > 0; i = i - 1)

{ for (int j = 1; j <= i; j = j + 1) cout << "*";


for (int j = 0; j <= loopsPerSecond/screenWidth; j = j + 1) ;

cout << "\n";

}

//Clear the screen

for (int i = 1; i <= screenHeight; i = i + 1)

{ for (int j = 0; j <= loopsPerSecond/screenWidth; j = j + 1) ;

cout << "\n";

}

return(0);

}

94 CHAPTER 5. LOOPS

5.4 Problems

1. Flowcharts revisited

Draw the flowchart representation (using lines, arrows, rectangles, diamond-shapes,and words) for the following looping constructs:

(a) the for-loop:for (i = 10; i > 0; i = i - 1){Sum = Sum + i;}

(b) the do/while loop:do{cin >> oddNumber; odd = oddNumber%2}while(!odd);

(c) the while loop:while (different){z = x; y = x; x = temp; different = x - y;}

2. Loop tests

(a) How many times is the following loop body executed?

int i = 0;

do

{ i = i + 1;

} while (i < 0);

(b) How many times is the following loop body executed?

int i = 0;

while(i < 0)

{ i = i + 1;

}

(c) How many times is the following loop executed?

int i = 3;

while(i)

{ i = i - 1;

}

(d) How many times is the following loop body executed?

int i = 0;

while(i <= 0)

{ i = i - 1;

}

(e) How many times is the following loop executed?

5.4. PROBLEMS 95

int j = 1;

do

{ j = -2*j + 1;

}while(j != 0);

(f) How many times is the following loop body executed?

for(int i = 0; i < 10; i = i + 2)

{ cout << "i = " << i << "\n";

}

3. Predict the output

The examples of code given below all compile correctly. Predict the output that thecode would produce if you compiled and ran it.

(a) int N = 0;

if (N = 0)

cout << "The conditional expression was TRUE\n";

else

cout << "The conditional expression was FALSE\n";

(b) int f;

for(f = 0; f <= 10; f = f + 2)

cout << f%10;

cout << "\n";

(c) int i = 0, Sum = 0, N = 3;

do

{ i = i + 1;

Sum = Sum + i;

}while(i <= N);

cout << "Sum = " << Sum << "\n";

(d) int i = 0;

if (0 != i)




(e) int j = 0;

if (0 < j);

{ j = j + 1;

}

cout << "j = " << j << "\n";

(f) int i = 2, j = -2;

while (j <= i)

96 CHAPTER 5. LOOPS

{ if (j < 0)

cout << "i is " << i << ".";

cout << " j is " << j << ".\n";

i = i - 1;

j = j + 1;

}

4. Loop drills

(a) Convert the following code containing a do/while loop to a code containing awhile loop in such a way that the execution of the code is identical.

int j = 0;

do

{ j = j + 1;

cout << j << "\n";

}while (j <= 10);

cout << j << "\n";

(b) Convert the following code containing a while loop to a code containing a do/whileloop in such a way that the execution of the code is identical.

int j = 10;

while (j > 0)

{ j = j - 2;

cout << j << "\n";

}

cout << j << "\n";

(c) Consider the following code:

#include <iostream>


int main(void)

{ int x = 1;

for (int i = 1; i <= 10 ; i = i + 1)

{ x = x*2;

cout << x << " ";

}

cout << endl;

return 0;

}

5.4. PROBLEMS 97

What does the above code print to the screen when you run it?

Draw the flow chart for the for loop in the above program.

(d) Consider the following code:

#include <iostream>


int main(void)

{ int i = 5;

do

{ i = i - 1;

cout <


int main(void)

{ int i = 13;

while(i%2)

{ i = i/3;

cout <


98 CHAPTER 5. LOOPS

int main(void)

{ for (int i = 1; i <= 2; i = i + 1)

for (int j = 1; j <= 3; j = j + 1)

cout <

#include <cmath>


int main(void)

{ float x;

cout << "Input x (must be positive): ";

cin >> x;

int i = 0;

const int NMax = 1000;

const float SMALL = 0.0001;

float xPower = 1.0, lastTerm, Sum = 0.0;

do

{ i = i + 1;

xPower = x*xPower;

lastTerm = xPower/i;

Sum = Sum + lastTerm;

}while (i < NMax && lastTerm > SMALL);

if (i == NMax)

cout << "Series did not converge\n";

else

{

cout << "i = " << i << "\n"

<< "x = " << x << "\n"

<< "lastTerm = " << lastTerm << "\n"

<< "Sum = " << Sum << "\n";

}

5.4. PROBLEMS 99

return 0;

}

(a) What series is being summed by this code? Sum = x + (you provide the rest)

(b) If the user inputs -1.0 for x, what will be printed out by this code?

(c) If the user inputs 0.0 for x, what will be printed out by this code?

(d) If the user inputs 1.0 for x, what will be printed out by this code?

(e) If the user inputs 0.0001 for x, what will be printed out by this code?

(f) If the user inputs 0.001 for x, what will be printed out by this code?

(g) If the user inputs 0.01 for x, what will be printed out by this code?

(h) If the user inputs 0.1 for x, what will be value of i that is printed out by thiscode? (A hand calculator may help.)

6. Think Like a Computer Again

Consider the following C++ code.

#include <iostream>

#include <cmath>


int main(void)

{ float x;

cout << "Enter x: ";

cin >> x;

int n = 0;

float lastTerm;

const float EPSILON = 0.001;

float sign = 1;

float denominator = 1;

float total = 1;

while(n <= 1 || lastTerm > EPSILON)

{ n = n + 1;

if (n/2*2 == n)

{ denominator = denominator*n*(n - 1);

lastTerm = pow(x,n)/denominator;

sign = -sign;

total = total + sign*lastTerm;

cout << "Term " << n << " is " << lastTerm << "\n";

100 CHAPTER 5. LOOPS

}

}

cout << "The total is " << total << "\n";

return 0;

}

(a) If you input 0.01 for x, what will be printed out?

(b) If you input 1.0 for x, what will be printed out? (A hand calculator may help.)

(c) What is another condition for the if statement that gives the same result?

(d) What is the formula this program is implementing?

7. Predict and convert

(a) Prediction part: If you compiled and ran this program, exactly what would theoutput look like?

#include <iostream>


int main(void)

{ const int maxTerms = 8;

const int x = 2;

int sign = 1;

int term = 1;

for (int i = 1; i <= maxTerms; i = i + 1)

{ sign = -sign;

term = term*x;

if (0 == i%2 || 2 <= i && 4 >= i)

{ cout < sign) cout << " - ";

else cout << " + ";

cout << term << endl;

}

}

return 0;

}

5.4. PROBLEMS 101

(b) Conversion part: Convert the program in the first part of this question to usea do-while loop rather than a for loop. Write your solution in the space below.

8. Predict and convert again

(a) Prediction part: If you compiled and ran this program, exactly what would theoutput look like?

#include <iostream>


int main(void)

{ const int maxTerms = 8;

const int x = 2;

int term = 1;

int sum = 1;

cout << "Sum = 1";

for (int i = 1; i <= maxTerms; i = i + 1)

{ term = -term*x;

if (0 != i%3 && 0 != i%4 )

{ if (0 > term) cout << " - " << -term;

else cout << " + " << term;

sum = sum + term;

}

}

cout << " = " << sum << "\n";

return 0;

}

(b) Conversion part: Convert the program in the first part of this question to usea while loop rather than a for loop. Write your solution in the space below.

9. Fix the bugs

Consider the following C++ code. It is an attempt to do the sum:

S = 1− x2 + x4 − x6 + x8 − x10

The code has errors that the compiler detects as indicated in the comment lines. Fixthem. The code also has “conceptual” errors in some lines, also indicated by comments.Conceptual errors are those that compile without error but do not provide correctanswers upon execution. Fix these as well.


#include <iostream>


int main(void)

{ float x;

cout << "Input x: ";

cin >> x;

float Sum, p; // Conceptual errors on this line

int N = 10;

do

{ i = i + 2; // Compiler detects error on this line

p = p*x; // Conceptual errors on this line

Sum = Sum + p;

}while(i <= N); // Compiler detects error on this line

cout << "Sum = " << Sum << "\n";

return 0;

}

10. More bugs to fix

Consider the following C++ code. It is an attempt to do the sum:

S = 1− x2 +x4

2!− x6

3!+

x8

4!− x10

5!

The code has errors that the compiler detects as indicated in the comment lines. Fixthem. The code also has “design” flaws in some lines, also indicated by comments.Design flaws are those that compile without error but do not provide correct answersupon execution. Fix these as well. The corrected code should be able to compile andcalculate the above sum.


int main(void)

{ cout >> "Input x:"; //Compiler detects error(s) on this line

cin << x; //Compiler error(s) on this line

double xPower = x; //Design flaw(s)

double sum; //Design flaw(s)

for(int term = 0; term <= 12; term = term + 1); //Design flaw(s)

xPower = xPower + x*term; //Design flaw(s)

5.4. PROBLEMS 103

sum = sum * xPower; //Design flaw(s)

cout << "Sum = " << Sum << "\n"; //Compiler error(s) on this line

return Sum; //Compiler detects error(s) on this line

}

11. Dealing cards

Complete the C++ code started below. The outer for loop deals 5 cards. Thestatement:int rank = rand()%13 + 1; chooses a random number between 1 and 13inclusive. So, a 1 is an ace, a 2 is a deuce, and so on up to 10, 11 is a jack, 12 is aqueen and 13 is a king. Then you must randomly pick the suit from one of 4: spades,clubs, diamonds and hearts. Your code must print out the results as in the example.Do not worry about having two cards the same as the deck your are dealing from is a6-deck superdeck, six full decks of 52 cards each, so dealing identical cards is possible,but not very likely.

Here is an example hand that was dealt by the program.:

Card 1 is a queen of spades.

Card 2 is a 3 of spades.

Card 3 is an ace of diamonds.

Card 4 is a king of diamonds.

Card 5 is a 7 of clubs.

#include <iostream>

#include <cstdlib>

#include <ctime>


int main(void)

{ srand(time(NULL));

for(int nCards = 1; nCards <= 5; nCards = nCards + 1)

{ int rank = rand()%13 + 1;

// You fill in the rest

.

.

.

.

} // End of the for loop over the 5 cards

return 0;

} // End of the main routine


12. Sentinel controlled input loops

Write a complete, standalone (if typed in, it would compile and run) C++ code thatprompts a user for an integer but only accepts it if it is a positive, odd integer less than100. If the user provides an integer that does not satisfy this, an explicit explanationis given and the user is prompted again. Your code has to provide output exactly asfollows. Computer output is in type font, user input is in italics.User! Type in an integer: 0The integer must be > 0. Try again.

User! Type in an integer: 100The integer must be < 100. Try again.

User! Type in an integer: 42The integer must be odd. Try again.

User! Type in an integer: 41Successful input!

13. Count the factors of 2

Write a C++ code that inputs a positive integer and prints out the number of factorsof two in it. For example, an odd number has no factors of two, 2 has one factor of2, 4 = 2 × 2 has 2 factors of 2, 6 = 3 × 2 has 1 factor of two, 8 = 2 × 2 × 2 has 3,10 = 5× 2 has 1 factor of 2, 12 = 3× 2× 2 has 2 factors of 2 and so on.

14. List the common factors

Write a C++ code that counts all the least common multipliers of a positive integerincluding 1 and the number itself. For example 8 = 1 × 2 × 2 × 2, 17 = 1 × 17,123456789 = 1× 3× 3× 3607× 3803. Here are sample uses of the program:

Input any positive int : 8

8 = 1 x 2 x 2 x 2


17 = 1 x 17


123456789 = 1 x 3 x 3 x 3607 x 3803

15. Compound interest

What happens if you deposit $1000.00 in a bank account that accumulates interest at5% per year, compounded annually? After the first year you have $1000.00× 1.05 =$1050.00 After the second year you will have $1050.00 × 1.05 = $1102.50. Write a

5.4. PROBLEMS 105

program that will accept a positive float for the starting balance and output how muchmoney you will have at the end of each year. The program stops when your moneyhas at least doubled.

Here’s an example of how it works:

% a.out

Input the starting balance: 1000

After year 1, balance = 1050

After year 2, balance = 1102.5














%

16. Oyster-Shucker’s Savings and Loan

Oyster-Shucker’s Savings and Loan has a great deal for you! This banks pays 1%interest per month, but only in months that do not have the letter “r” in its name (arather strange but lucrative deal). Thus, for each month with an “r”, the total amountof money is multiplied by 1.01. For months without an “r”, the total is unchanged.

Write a program that:

• Reads in an int representing the first month the money is put in the bank,

• Reads in an int representing the number of months the money is in the bank,

• Reads in a float for the amount of money deposited,

• Loops over the number of months that the money is in the bank, and,

– If the current month has an “r”, multiplies the total by 1.01,

• And finally, prints out the final amount of money.

You may find the following useful: The statement


currentMonth = (startMonth + i) % 12;

calculates the month, given a starting month and a number of months elapsed. Forexample, if you start in January (Month 1), 13 months later it is February (Month 2).

17. There’s Gold in Them Thar Ratios

In the Fibonacci series1, 1, 2, 3, 5, 8, . . .

each number (after the first two) is equal to the sum of the two before it. So, forexample, the next number after 8 is 8 + 5 = 13. One of the properties of the series isthat the ratio of a number in the series to the one immediately preceding it converges(when the Fibonacci get very large) to the so-called Golden Ratio.

Write a program to approximate the Golden Ratio. Your program should computeterms in the Fibonacci series, taking the ratio of the most recent term to the onebefore it. Once the ratio is not appreciably different from the last time it was computed,terminate the loop and print out the result.

18. The Mersenne primes

A “prime number” is a positive whole number that can only be divided by itself and 1to produce another positive whole number. The first few primes are 1,2,3,5,7,11... InC++ language the test:if (0 == N%i)

would never be true for a candidate prime N if i were greater than 1 and less than N.

A Mersenne prime is a prime number that also has the form 2p−1 where p is a positivenumber greater than 1.

Write a C++ code that calculates candidate Mersenne primes 2p − 1 for 2 ≤ p ≤ 31.(This step is coded for you below.) Then test the number to see if it is prime. (You’llneed a loop to do this.) If you detect that the number is prime, print out the numberand the value of p.

19. Nest your for’s, inside and out

Write a C++ program that will:

(a) Prompt the user for a positive odd integer within a do-while construct. If theinteger is 0, negative or even, prompt the user again. The algorithm goes to thenext step when an odd positive integer is received. Call this odd integer N.

(b) Output an inverted triangle of asterisks that looks like the following example.The first row has N *’s. Each following row has two *’s less than the one above itand is centered with respect to the row above.

Here’s an example of how the program should work:

5.4. PROBLEMS 107

red% a.out

Input an odd positive int: -1

Input an odd positive int: 0



***********

*********

*******

*****

***

*

red%

You must use a do{}while(); loop to obtain N.

You must use for-loops to make the drawing.

You may not use arrays.

You must only use cout statements that have only single character strings, for ex-ample, cout << "*";. Something like cout << "**"; or cout << " *"; is notpermitted. Spaces are counted as a character.

Your program must work for arbitrary odd, positive N.

20. Nest your for’s, left to right

Write a C++ program that will output the following:

****************************************

*********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *


* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ********** *

* ***********

* **********

* *********

* ********

* *******

* ******

* *****

* ****

* ***

* **

****************************************

You ...

must use for-loops, some nested, some not.

may not use arrays. (Ignore this statement of you do not know what an array is.)

must only use cout statements that have only single character strings, for example,cout << "*";. Something like cout << "**"; or cout << " *"; is not permit-ted. Spaces are counted as a character. For example, the top line would madecorrectly as follows:

const int width = 4;

for (int i = 1; i <= width; i = i + 1) cout << "*";

cout << endl;

rather than the following which is not allowed,

cout << "**************************************************";

The width and length of the drawing are exactly 40 characters.

5.4. PROBLEMS 109

21. For success, nest your for’s

In the output that follows, a C++ program prompted a user for two inputs, the heightand width of a box and the user responded with “30” for both. Write the C++ programthat did this:

Height of box: 30

Width of box: 30

* * * * * * * * * * * * * * *

** ** ** ** ** ** ** ** ** **

*** *** *** *** *** *** *** **

**** **** **** **** **** ****

***** ***** ***** ***** *****

****** ****** ****** ****** **

******* ******* ******* ******

******** ******** ******** ***

********* ********* *********

********** ********** ********

*********** *********** ******

************ ************ ****

************* ************* **

************** **************

*************** **************

**************** *************

***************** ************

****************** ***********

******************* **********

******************** *********

********************* ********

********************** *******

*********************** ******

************************ *****

************************* ****

************************** ***

*************************** **

**************************** *

*****************************

You must use nested for-loops to solve this problem.

You may not use arrays. (Ignore this statement if you do not know what an arrayis.)


You must only use cout statements that have only single character, for example,cout << "*";. Something like cout << "**"; or cout << " *"; is not permit-ted. Spaces are counted as a character.

22. Base Pythagorean Triples

A base Pythagorean triple is defined as three integers, i, j, k such that i2+ j2 = k2 andi, j, k have no common divisors other than 1. For example 3, 4, 5 is a base Pythagoreantriple. However, 6, 8, 10, although 62+82 = 102, is not a base Pythagorean triple since6, 8, 10 are all divisible by 2. Write a program that finds all the base Pythagoreantriples such that i, j, k ≤ 1000.

Here is an example of running the code:

Maximum integer to test = 1000

Triple # 1: 3*3 + 4*4 = 5*5 = 25

Triple # 2: 5*5 + 12*12 = 13*13 = 169

Triple # 3: 8*8 + 15*15 = 17*17 = 289

Triple # 4: 7*7 + 24*24 = 25*25 = 625

Triple # 5: 20*20 + 21*21 = 29*29 = 841

Triple # 6: 12*12 + 35*35 = 37*37 = 1369

Triple # 7: 9*9 + 40*40 = 41*41 = 1681

Triple # 8: 28*28 + 45*45 = 53*53 = 2809

Triple # 9: 11*11 + 60*60 = 61*61 = 3721

Triple # 10: 16*16 + 63*63 = 65*65 = 4225

Triple # 11: 33*33 + 56*56 = 65*65 = 4225

Triple # 12: 48*48 + 55*55 = 73*73 = 5329

Triple # 13: 13*13 + 84*84 = 85*85 = 7225

Triple # 14: 36*36 + 77*77 = 85*85 = 7225

Triple # 15: 39*39 + 80*80 = 89*89 = 7921

Triple # 16: 65*65 + 72*72 = 97*97 = 9409

Triple # 17: 20*20 + 99*99 = 101*101 = 10201

.

.

.

5.5. PROJECTS 111

5.5 Projects

1. Something loopy to do

Write a C++ program that converts any unsigned integer to any base between 2 and9.


unix-prompt> a.out

Input the base to convert to between 2 <= b <= 9: 7

Input any positive unsigned int: 126


unix-prompt>

2. Going nowhere fast

In this problem the program starts up assuming that it is following an object on a2-dimensional plane as it takes many steps to perform a sort-of “random walk”. Theobject starts off at the “origin” (x = 0 and y = 0). You will input two real numbersthat will represent the next coordinates of the object. The computer will calculatethe straight-line distance that it has to travel to get there and position the object atthat point. Then you will input two more real numbers representing a move from thatlocation, then another position (another pair of points) and another position (pair ofpoints) and so on. To stop the program you have to input the identical new coordinatesas the previous step—that is, there is no distance to be traveled for that final step.This is a “sentinel” indicating that the loop should terminate. This last zero-stepshould not be counted in the number of steps. When the program exits the loop atthis point make the program print out the total distance traveled, the total number ofmoves and the average distance per move.

Here is some pseudocode for this problem:

(a) START: Declare N = 0 (Counter for the number of steps), xi, yi (New coordi-nates), x0 = 0, y0 = 0 (Current coordinates), s (Distance of a single step) andt = 0 (Total distance).

(b) Start of the “input/stepping loop”.

i. Read in the the new coordinates: xi and yi.

ii. Test to see if xi and yi are valid input. If they are not, go back to (b-i.),otherwise continue on to the next step (b-iii.).


iii. Test to see if xi and yi are identical to x0 and y0. If they are not you willcontinue with step (b-iv.) Otherwise you will want to transfer control to thethe first statement following the “input/stepping loop”, step (c).

iv. Increment the step counter N .

v. Calculate the distance of the step s =√(xi − x0)2 + (yi − y0)2.

vi. Accumulate s in t, the total distance.

vii. Important! Update the current position x0 = xi, y0 = yi.

viii. Try to take another step, that is return control to the top of the “input/steppingloop”, step (b-i.).

(c) Report the total distance traveled.

(d) Report the total number of steps.

(e) Report the average distance traveled per step. Important: make sure that thecase of zero steps and zero distance traveled does not cause problems.

(f) END:

5.5. PROJECTS 113

3. The exponential function

(a) Computation of the factorial

Consider the following C++ program:

#include <iostream>


int main(void)

{ cout << "Input N: "; int N; cin >> N;

if (0 > N)

cout << "N < 0. Stopping.\n";

else if (0 == N)

cout << "0! = 1\n";

else

{ int factorial = 1;

for (int i = 1 ; i <= N; i = i +1)

factorial = factorial*i;

cout << N << "! = " << factorial << "\n";

}

return 0;

}

This program computes the “factorial” of the integer N, input by the user. Thefactorial of N , represented by the mathematical symbol N !, is calculated by:

N ! = N × (N − 1)× (N − 2)× (N − 3) · · ·3× 2× 1

By definition, 0! = 1. Other factorials are 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120and so on. Factorials are used a lot in many branches of mathematics.

Your problem is to find out for what values of N input by the user, does theprogram above not give correct answers. (Here’s a hint: The maximum signedinteger has a value 231 − 1 or 2147483647.) The way you determine this I leaveup to you. You can do it mathematically, or you can run the above programfor different values of N until you detect that something is wrong. Turn in adescription of how you found the answer.

(b) An improved computation of the factorial

Convert the above program so that the variable called factorial is a float.(Turn in your code.) Now, try again to find out for what values of N input bythe user, does the program above not give correct answers. (Here’s another hint:The maximum float has a value 3.40282347 × 1038.) You may determine this


mathematically, or you can run your program for different values of N until youdetect that something is wrong. How do you know that something has gonewrong? Turn in a description of how you found the answer.

(c) Calculating the exponential of a number

The exponential function, called ex is one of the most common functions in all ofmathematics. One way of calculating it is to use the following:

ex =1

0!+

x

1!+

x2

2!+

x3

3!+

x4

4!+

x5

5!· · ·+ xn

n!· · ·

an infinite series, one that never stops. Mathematically, the above expression forex is exact for any x from minus infinity to plus infinity. However, to get anaccurate calculation from a computer requires some work. Here is what you haveto do:

i. Write a C++ program to compute ex. Your program should try to computeit by summing the finite series

ex =1

0!+

x

1!+

x2

2!+

x3

3!+

x4

4!+

x5

5!· · ·+ xN

N !

where N is some integer that you, the programmer, will set. Be carefulthat you do not set N too high, otherwise, as indicated in the previous twoproblems, your answer may not make sense. N should also be great enoughso that your answer is as accurate as it can be. Turn in your code.

ii. Discuss over what range of x you expect your program to provide a reasonablyaccurate answer. Turn in your discussion in such a way that pleases your LabInstructor.

You can check your results with a calculator. Else, if you read ahead, you cancheck your program using the C++ math function exp(x).

5.5. PROJECTS 115

4. Primal loop therapy

This question is all about loops. A solution for this may involve all of the loops wehave encountered: the do{}while() loop, the for(){} loop, and the while(){} loop.A solution also facilitated with the use of the modulus operator (%).

A prime number is a positive whole number that can be divided perfectly only by itselfand one. The first 26 prime numbers are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97. You will write a C++ code thatwill accept two integers from its user. You will restrict the range of inputs to between4 and 1,000,000 (one million). The user does not know this, he or she only knows toinput integers and so you will keep prompting the user until he or she gives you properinput. (Sentinel-controlled do{}while() loops are best for this job.) The two integerscan be accepted in any order, bigger one first or smaller one first. You can assume thatthe user is not “hostile”. The user will only input integers, not floats nor charactersnor anything else that cin might find disturbing when asking for an int.

Once you have two integers you can work with, you will find all the prime numbersbetween those two numbers, including the starting and ending numbers themselves.You will output the prime numbers as you find them to the screen, and as a last step,print out how many of them you found.

Hint 1: Make sure that the primes your program gives agree with the ones in the listabove.Hint 2: There are 78,496 prime numbers between 4 and one million.

Hint 3: If you are using Hint 2 as a way of checking your program (please do!) andyou find that it takes a very, very long time to get your program to complete its task,stop a moment and think! Think about how many potential divisors you have to testto see if a number is prime or not. It should not take more than about a minute orso to find all the primes between 4 and one million, depending on what machine youare using. On a 700 MHz PIII machine a solution takes about 70 seconds if it outputsevery number it finds to the screen and only about 10 seconds if it outputs only thenumber of primes found. To be sure, there are even faster ways of doing it!

Finding prime numbers is an exercise that has intrigued mathematicians for centuries.As soon as computers were invented, this was one of the early problems that mathemati-cians programmed them for. We know that integers have only 32-bit representations.This is just food for thought, not part of the question: How you would find out if anumber bigger than 232 is prime? Now that’s a tough computer problem. Here is anexample of output from correct solution:

> g++ primes.cpp

> a.out

Input the first integer: 1000001


Input must be > 4 or < 1000000

Keep trying!

Input the first integer: 1000000

Input the second integer: 1

Input must be > 4 or < 1000000

Keep trying!

Input the second integer: 4

Looking for all the prime numbers between 4 and 1000000 inclusive

5 is prime

7 is prime

11 is prime

13 is prime

17 is prime

19 is prime

23 is prime

29 is prime

31 is prime

37 is prime

41 is prime

43 is prime

47 is prime

53 is prime

59 is prime

61 is prime

67 is prime

71 is prime

73 is prime

79 is prime

83 is prime

89 is prime

97 is prime

.

.

.

999953 is prime

999959 is prime

999961 is prime

999979 is prime

999983 is prime

There were 78496 prime numbers between 4 and 1000000 inclusive

5.5. PROJECTS 117

5. Where’s Waldo?

It is a little known fact that Waldo Schtinklebaum, the character of the infamous“Where’s Waldo?” game was a student here at the University Michigan. He startedoff his career in Civil Engineering. He was notorious for 1) being a lousy dresser, 2)having no sense of direction (why he was always getting lost), 3) and being indecisive.In fact, after partying all night at the Frats, Waldo would often lose his way backto his dorm. He’d leave the party, forget which way his dorm was, take a step in arandom direction, change his mind and his direction and take another step. This is nota bad approach to getting where you want to go, except that it can take a very longtime to reach your destination. Waldo sometimes ended up walking around randomlyuntil the morning when someone would take pity on him and lead him back home. Infact, Waldo’s meanderings led to his second career. Shortly after graduation, Waldoparticipated in the infamous Spring running event that takes place in the late eveningon the last day of Winter term. On this day, Waldo changed his manner of dress—permanently. Despite his motivation to run in the correct direction, Waldo ended upin the Kresge Eye Center were his vision was corrected and he donned the hospitalgreens of a medical doctor. Waldo is now an optometrist working in Livonia under anassumed name. His specialty is treating children who have gone cross-eyed staring atthe puzzles that bears his former name.

Your project is to write a program that imitates Waldo’s after-party mission to gethome. A pseudocode description of the solution is given below.

Pseudocode for Waldo’s mission

• Waldo starts his journey at location (x, y) = (0, 0), the front door of the Frathouse, the center of the universe that Waldo is in. Waldo’s universe is a 2-dimensional plane that measures 20×20, that is −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10.Waldo’s home is a 2× 2 square in the upper right hand corner, 8 ≤ x ≤ 10 and8 ≤ y ≤ 10.

• Declare a variable called stillWalking and give it an initial value of 1 (true),which means that Waldo is still trying to find his way home on foot.

• Declare another variable called atHome and give it an initial value of 0 (false)which means that Waldo has not yet found his way home.

• Declare another variable called stepsToTry and give it an initial fixed value of1000. Waldo hopes to get home before 1000 steps. If he doesn’t, you will eithertell him to keep going and try another 1000 steps, or to give up and sleep on thenearest park bench.

• WHILE (stillWalking)

– FOR-LOOP until 1000 steps are taken and while Waldo has not yet reachedhome.

∗ DO (Loop for choosing the direction of the step)


· Take a proposed step of unit length in a random direction. (See noteon the next page.)

WHILE (endpoint of the step is outside of Waldo’s universe)

∗ Take the step.

∗ If Waldo’s new location is within the square defined as his home, setatHome to true.

END FOR-LOOP

– IF (atHome), print out the total number of steps that Waldo has taken andstop the program.

– ELSE, read in a new value for stillWalking. If the user inputs a non-zerovalue for stillWalking, Waldo will try to take another 1000 steps, otherwise,if the user inputs a zero value for stillWalking the logical condition of theouter WHILE loop will be false.

END WHILE

• End the program.

Note 1: Taking a step in a random direction

To use random numbers, you must

#include <cstdlib>

in the preprocessor area for the code. Then, an expression of the form:

rand()

will produce a random int between 0 and RAND MAX inclusive, where RAND MAX is a bigpositive number defined in cstdlib. This random int must be converted to a randomdouble that lies between 0 and 2π. This can be done via the following statement:

double angle = 2*PI*rand()/RAND_MAX;

where PI can be declared and initialized (earlier in the program) by

const double PI = 4*atan(1.0);

Using the above atan() function requires the use of

#include <cmath>

in the preprocessor area. Finally, x and y displacements can be generated by thestatements:

5.5. PROJECTS 119

dx = cos(angle);

dy = sin(angle);

where dx and dy are double’s. This is one of Waldo’s steps. You should use exactlythis order of displacement statements in your solution.

Example use of the program

red% a.out

Will Waldo make it home? Press <RETURN> to continue.

Waldo is still lost after taking 1000 steps

Continue playing? (Input "0" to stop): 1

Waldo is still lost after taking 2000 steps

Continue playing? (Input "0" to stop): 1

Waldo made it home after taking 2552 steps!

Note 2: Making it really sort of random (optional topic)

If you have followed the above instructions carefully, you will notice that every exe-cution of the program is exactly the same. Moreover, the solution of your colleagues,at least those who followed the instructions carefully, will also get Waldo home in thesame number of steps (assuming you all are using the same type of computer). In thiscase, it is a desirable thing because, it allows your GSI’s to grade your assignmentquickly. This makes your GSI’s happy because most GSI’s would volunteer to undergoa frontal lobotomy rather than grade. (I used to be a GSI and I know that I would.)Happy GSI’s give out better grades, it’s a fact of life. I’m digressing again...

If you want to make each play of the game different, do the following. In the prepro-cessor area of your code,

#include <ctime>

Then, in your code before your first use of rand(), put in the following expression:

srand(time(NULL));

That’s how I generated the different solutions that are shown in graphical form on thenext page. The expression time(NULL) gives the number of seconds since the start ofthe epoch, January 1, 1970, 12:00am GMT. This is considered to be the start of themodern computer age.

Note that you can also “seed” the random number generator rand() using the followingstatement:


srand(someInt);

where someInt is an int that you can choose. If you choose the same someInt everytime, the game will be the same. Different someInt’s result in different games. Try it!

Some things to think about (optional)

These are some things to think about. Don’t hand answers to these to your GSI’s.However, feel free to speak to them or me (the Prof) about it.

• Why did I confine Waldo to a 20× 20 universe?

• Is Waldo always guaranteed to make it home if he tried long enough?

• If I took away the walls of the universe, on average how far away would Waldo befrom his starting point after a certain number of steps.

• If you took away the walls, would Waldo always guaranteed to make it home ifhe tried long enough?

Graphical output of Waldo’s meanderings on next page

Waldo is there. Can you see him?

5.5. PROJECTS 121

Wandering WaldoFirst example

Waldo’s tortuous path

Waldo starts here

HomeWaldogetsinspired!

Wandering WaldoSecond example


Chapter 6

Early Abstraction—Functions

6.1 Motivation for functions

Let us consider an implementation of the al-Kashi algorithm. This algorithm computes theXN where N is any integer, positive, negative or non-zero. Here is what the C++-codelooks like (This code is called alKashi.cpp on the lecture distribution area on the coursewebsite):

//File: alKashi.cpp

#include <iostream>


int main(void)

{ cout << "A calculation of X (float) to the power of N (int) \n";

float X; // The float to be raised to a power

int N; // Power to be raised to

cout << "Input X (float) and N (int): ";

cin >> X >> N;

float result; // The result of the calculation

{ //Open a new "scope" or "code" block

//An implementation of the al-Kashi algorithm

int n = N; //An int used internally in this code block

float x = X; //A float used internally in this code block

123

124 CHAPTER 6. EARLY ABSTRACTION—FUNCTIONS

if (0 > N)

{ //This "trick" allows us to calculate for negative powers!

n = -N;

x = 1/X;

}

result = 1; // Initial value

while(1 <= n)

{ if (0 == n % 2)

{ n = n/2;

x = x*x;

}

else

{ n = n - 1;

result = result*x;

}

}

} // End of the scope block

cout << X << " to the power " << N << " = " << result << "\n";

return(0);

}

The above example introduces a new concept in C++-programming, that of a “block”, or“code block”. A “block” is any set of statements enclosed by a set of curly brackets {...}’s.In the above example,


.

.

.


delimit a code block—this one associated with the Al-Kashi algorithm. Variables can bedefined within a code block. Those that are defined within them (in this example, float x

and int n) are not known (or defined) outside of the the block. If you tried to do somethingwith x or n outside of their block, you would get a compilation error, complaining that thesevariables were not defined. At least, that will be our understanding of code blocks until weget to the section called “scope rules” a little later in the course. We have actually seencode blocks before. The main routine has one, as well as the if/else if/else construct,the while and do/while constructs as well as the for loop.

6.1. MOTIVATION FOR FUNCTIONS 125

Now, suppose that we did not know about the al-Kashi algorithm and coded the algorithm ina more straightforward way (This code is called dumbPower.cpp on the lecture distributionarea on the course website.):

//File: dumbPower.cpp

#include <iostream>


int main(void)





cin >> X >> N;



//An implementation of the dumbPower algorithm

int n = N;

float x = X;

if (0 > N)

{ n = -N;

x = 1/X;

}

result = 1; // Initial value

for (int i = 1 ; i <= n ; i = i + 1)

{ result = result*x;

}



return(0);

}

In these two examples, we note that there is a lot of repetition in the code outside of thescope block. Repetitious coding is BAD! It means that someone has wasted a lot of time!


Indeed, as far as the main routine is concerned, it does not really care about the contents ofthe code block! All it really wants is the answer. It can think of the algorithm to computethe power in a very abstract way. All main really needs to do is to give the scope block thevariables X and N.

What we really, REALLY want is a routine that is independent of the specific calculationaltechnique that is used to obtain XN . In fact there is. There is a “function” in the C++standard math library that does this. It is called pow. If we use this information we canwrite a much more compact code (This code is called power0.cpp on the lecture distributionarea on the course website.):

//File: power0.cpp

#include <iostream>

#include <cmath>


int main(void)





cin >> X >> N;


result = pow(X,N);


return(0);

}

Note the use of

#include <cmath>

that helps with the interpretation of the statement:

result = pow(X,N);

6.1. MOTIVATION FOR FUNCTIONS 127

This is a much more abstract and powerful implementation of the previous two examples.It allows the main routine to be concerned with only the input and output of the variousparameters, X, N, and the answer. The main routine in this case is much more general.

Let us pretend that we are now in the infancy of the C++-language, and that the pow functiondoes not exist. However, we require this functionality repeatedly in our applications. But,we do know about al-Kashi’s algorithm. Wanting to be efficient, we would code the al-Kashialgorithm into a C++-function and employ it as follows (This code is called power1.cpp onthe classcodes distribution area):

//File: power1.cpp

#include <iostream>


float pow(float x, int n)

{ float result; // The result of the calculation


if (0 > n)

{ n = -n;

x = 1/x;

}

result = 1;

while(1 <= n)

{ if (0 == n % 2)

{ n = n/2;

x = x*x;

}

else

{ n = n - 1;

result = result*x;

}

}

return result;

}

int main(void)






cin >> X >> N;


result = pow(X,N);


return(0);

}

Now we note a few important things. There are a lot of things to remember about functionsand we are only going to scratch the surface.

• In our use of functions, the function definitions and the main routine go in one file.(They can be put into different files but it becomes more complicated.)

• The function definitions go at the top of the file, after the pre-processor directives butbefore the definition of the main routine. (They can be put after the main routine butit becomes more complicated.)

• The functions definitions should go before they are used in any other function, includingmain. (Again, we can use another ordering, but it becomes more complicated.)

• Functions, the way we have used them so far, can return only one thing via the state-ment

return expression.

• The expression in the return line must evaluate to a data type that is specified inthe definition of the function, in this example, float pow()...

• The parameters in the function definition, in this example, (float x, int n), mustbe declared.

• The variables in the function call, in this example, X,N from main’s function callresult = pow(X,N);

are not changed by the function. (There is another way to call functions where theparameters can change. This will be taught later.)

6.2. USER-DEFINED FUNCTIONS 129

• Once the function returns to its calling routine, main in this case, the variables inthe function’s parameter list and any other variables that it declares, are lost. Theonly thing that main has is the return value. (To be absolutely truthful, there areways that the calling routine can look at the memory where the function stored thingstemporarily, but...it’s complicated! Moreover, it is almost never done.)

• Functions can call other functions.

• main is just another function.

• If a function does not return anything, it’s type is void. Example,

void returnNothingFunction(void){}

• If a function has no parameters, put void where the parameter list goes, as in theabove example.

• Finally, functions are really, really, REALLY important. So it is important to un-derstand them and use them well. We will spend a lot of effort in this course withfunctions.

6.2 User-defined functions

In this section, we examine the syntax associated with functions.

There are three things that are associated with the definition of a function in ANSI C++:

• The function prototype: For small applications, where all the source code is found inone file, the function prototype is usually written in the pre-processor area, the areabefore any function definition (see below). In small applications, where all the sourcecode is found in one file, the function prototype can be omitted if the function defini-tion is located in the file before any function call (see below) is made to it. For largeapplications, or applications that have more than one source file, function prototypesare usually gathered into an “include file”, named, for example, myIncludes.h and ac-cessed via an include statement #include <myIncludes.h> placed in the preprocessorarea.

• The function definition: For small applications, where all the source code is found inone file, the function definition is written after the function prototype and outsideof the main routine. For large applications, or applications that have more than onesource file, it is conventional to have one function definition per source file.


• The function call: The call to the function can be inside of main or another function.In fact, a function can even call itself. This is called “recursion” and is discussed in alater chapter.

Understanding the syntax of these three things is crucial to the successful use of functions.

The general use of a user-defined function is as follows:

/* Opening comments */

.

.

.

// Start of pre-processor directives

.

.

.

// function prototype (if required)

(type) myFunction(variable type list separated by ,’s);

.

.

.

// end of pre-processor directives

.

.

.

/* function definition */

(type) myFunction(variable type list and variables separated by ,’s)

{

.

.

.

/* Function body with declarations and

executable statements

*/

.

.

.

}

.

.

.

int main (void)

{ // Start of main routine


.

.

.

// function usage

...myFunction(list of values separated by ,’s)...;

.

.

.

} // End of main routine

General comments on functions:

• The function prototype is the declaration of a function. It specifies what the functionis expected to provide (return value, see below), what is in the parameter list (insidethe parentheses) and its identifier (name). You do not need a function prototype ifall your source code is located in one file and if the first time the compiler encountersyour function within a source code file is the function definition itself. The compilerwill form its own prototype from the definition! This is why we can get away withnot using prototypes in most cases when everything is in the same file, the functiondefinitions are in the proper order and main comes last. There are exceptions to thisrule but we will probably not encounter them in this course. If the compiler encountersthe function for the first time in the function call, it will form its own prototype andassume that the return type is an int and can potentially mess up on the argumentlist as well. Avoid this situation at all costs!

• The (type) in the “function prototype” and “function definition” inform the compilerthat the function called myFunction will return a variable of a certain type. (type)

could be float, e.g.

float myFunction();

which means that myFunction is expected to return a float.

• Other possibilities for (type) could be int or any other variable type in C++ (thathave not been discussed yet in this course).

• If (type) is not defined, then the compiler assumes that it is an int.

• The other possibility for (type) is void. void means that the function will not returna value. (It will perform a function on some data but not return any kind of value.)

• (type) in the “function prototype” and “function definition” must agree, else therewill be a syntax error.


• Forgetting to return a value when (type) indicates that one should be returned resultsin unpredictable behavior.

• Returning a value when (type) is void causes a syntax error.

• Note that

int main (void)

{

.

.

.

}

is a function definition! Its (type) is int. We could have said

main ()

{ .

.

.

}

because functions are assumed to be (type) int unless we state otherwise. main is aspecial function in C++ that the operating system knows is the place where to startprocessing. Only main has this special status. I always (type) every function that Idefine and you should do the same.

• The rules for naming a function, e.g. myFunction, are the same rules as for variables,i.e. the normal identifier naming rules (≤ 31 alphanumeric characters and the under-score character “ ” and NOT starting with a number.)

• The argument list in the function prototype is a comma separated list of variable types,e.g.

void myFunction(float, int, float);

which indicates myFunction will be called and defined with a float, an int, and afloat in that order. Here’s another example:

void realQuadraticRoots(float, float, float);

A variable type is required for any non-int. Therefore, declaring a variable position asan int is optional. However, it is recommended that you declare the variable type forall members in the parameter list.


• The parameter list can also be void.

• The parameter list in the function definition follows the same rules except that thevariables must be named. e.g.

void realQuadraticRoots(float a, float b, float c){/* Function definition */}

• The function call (within the body of the main routine or within the body of anotherfunction) as shown here:

int main (void)

{

.

.

.

realQuadraticRoots(a, b, c);

.

.

.

}

transmits the values of the variables in the parameter list to the function. This istermed “call by value”. There is another way of doing this termed “call by reference”which we will discuss later.

• The variables defined in the function definition (as well as any other variables that thefunction defines) are internal to that function, even if the name (identifier) is identicalto a name elsewhere in the program.

• Functions must all be defined in a separate code segment. You can not define a functionwithin another function.

• Function prototypes outside of the body of any function apply to all function calls inthat source file.

• Function prototypes inside the body of any function apply only to the associated callswithin that function.

• A function can return in 3 ways:

1. by running into its closing brace } (this will work only for void functions)

2. running into the following statement anywhere in the function body

return;


This will work only for void functions.

3. running into the following statement anywhere in the function body

return (an expression);

This is the way non-void functions must return. The parentheses around “anexpression” are optional. The expression should evaluate to the same variabletype that the function is expected to return.

• Functions can be created independently by different programmers. The only infor-mation that has to be agreed upon is the return type, the function names and theparameter lists. Otherwise, each programmer can do his or her job however he or shepleases. In a team setting, prototypes are usually decided upon at the design stage.An example follows.

6.3 Example: A problem tackled with teamwork

(Completed version posted on the ENG101 web site in the lectures area called quadraticRoots.cpp)

A team of 4 people, Prof (the boss) and his graduate 3 students, Bill, Jane and Eugene (themath expert) are writing a program together.

• Prof to Team: Team: We will write a code to solve the quadratic equation! We’ll patentit and I’ll get rich!

• Team to Prof: Yeah, right!

• Prof to Bill: Bill, I want you to write a bool function called realQuadraticRoots thatwill return an integer, and accept the three floats a, b and c in that order. If there areno real roots your function has to return the value false. If there are, it has to returnthe value true.

• Bill to Prof: Um, Prof, what’s a bool??

• Prof to Bill: It’s a variable type that can only have two values, true or false. It isdeclared and assigned value as follows:

bool truthOrDare = false;

• Bill to Prof: Gotcha, I’m on it. By the way, what’s my cut?

• Prof to Jane: Jane, I want you to write a function called numberQuadraticRoots thatwill return an integer, and accept the three floats a, b and c in that order. I will onlycall your routine if Bill’s routine tells me that there are real roots. The integer youreturn should contain the number of real roots, either “1” or “2”.

6.3. EXAMPLE: A PROBLEM TACKLED WITH TEAMWORK 135

• Jane to Prof: OK, it’s clear what I have to do.

• Prof to Eugene: Eugene, you do the math! I want you to...

• Eugene to Prof: You probably want me to write a function called something liketwoQuadraticRoots that will return no value, and accept the three floats a, b andc in that order and print a statement telling you what the two roots are. You will onlycall your routine if Jane’s routine tells you that there are two real roots. Right?

• Prof to Eugene: Eugene, stop interrupting! I want you to write a function calledtwoQuadraticRoots that will return no value, and accept the three floats a, b and c

in that order and print a statement telling me what the two roots are. I will only callyour routine if Jane’s routine tells me that there are two real roots.

• Prof to Team: Get to work!

• Team to Prof: What’s in it for us?

• Prof to Team: Stop kvetching already! Here’s the main routine with function stubs.You guys fill in the blanks. My code compiles so it’s absolutely perfect.

Here’s the Prof’s main routine with “stub” functions:

//File: quadraticRootsStubs.cpp

#include <iostream>

//Do I need anything else, fellow programmers?


//Function stubs

bool realQuadraticRoots(float a, float b, float c)

{

}

int numberQuadraticRoots(float a,float b,float c)

{

}

void twoQuadraticRoots(float A, float B, float C)

{

}

int main(void)

{ cout << "a, b, c: ";

float a, b, c;


cin >> a >> b >> c;

if (realQuadraticRoots(a,b,c) && 2 == numberQuadraticRoots(a,b,c))

twoQuadraticRoots(a, b, c);

return 0;

}

Here is what the team produced.

Here’s Bill’s code. Note that the Prof did not fully specify his task and so he let him haveit. He had to find a way to terminate the program from a subroutine and so he had to go tothe library and do some research. (The Prof never answers e-mail from graduate students!)Bill learned how to do this with the exit function and other definitions given in cstdlib.

bool realQuadraticRoots(float a, float b, float c)

/*********************************************************************

Purpose: See if there are real roots

Receives: Quadratic constants a,b,c in a*x*x + b*x + c = 0

Returns: true if real roots, false if none

Programmer: Bill Boole (with attitude)

Start Date: 09/17/00

Remark1: Needs <iostream>, <cstdlib>

Remark2: Prof needs to rethink this thing

*********************************************************************/

{ if (0 > (b*b - 4*a*c)) return false;

else if (0 == a && 0 == b)

{ cout << "Prof, you numbskull!\n"

<< "You didn’t tell me what to do if a and b are both 0!\n"

<< "Go back and redesign the code!\n";

exit(EXIT_FAILURE);

}

else return true;

}

Here’s Jane’s code. Her job was made easier by Bill’s code letting her know that there is atleast one solution.

int numberQuadraticRoots(float a,float b,float c)

/*********************************************************************


6.3. EXAMPLE: A PROBLEM TACKLED WITH TEAMWORK 137


Returns: 1 if one real root, 2 if two real roots

Programmer: Jane Justintime


Remarks: Enters this routine knowing that there are roots

*********************************************************************/

{ if (0 == a || (0 == (b*b - 4*a*c))) return 1;

else return 2;

}

Eugene, who thinks he’s a genius and a super-geek, likes to do things his own way. Helikes to store all the values his function receives in uppercase variables. Eugene has his owncoding style. Eugene also thinks he is smarter than the compiler and so he played with thequadratic equation to put it into a form that he thinks will execute faster. (You can verifyfor yourselves that what Eugene did was correct mathematically.)

void twoQuadraticRoots(float A, float B, float C)

/*********************************************************************



Returns: The 2 two real roots of the quadratic equation

Programmer: Eugene Dweebowski (the genius)


Remark1: Needs <cmath>

Remark2: Enters knowing that there are two real roots

Remark3: I dare ya to find something faster and better!

*********************************************************************/

{ float twoA = 2*A, twoC = 2*C, radical = sqrt(B*B - twoA*twoC) - B;

cout << "Solution 1: " << radical/twoA << "\n";

cout << "Solution 2: " << twoC/radical << "\n";

}

Some comments are in order for the Prof’s code. It is incomplete. It does not do anythingwhen there is only one solution. His code needs a little refinement before it can be consideredcomplete. And, he has to decide what to do if all the constants are zero.

As a postscript to this story, function stubs are a good alternative to function prototypesfor small projects. Note that the Prof’s stub code will compile so that he can at least checkhis programming. However, for larger projects, prototypes are the way to go.


6.4 Call by value, call by reference, reference parame-

ters

In our present usage of function calls, the parameter lists have contained only values thatare transmitted to the function and stored in internal variables known only to that function.This is given the name call by value. A function that is called “by value” operates on thedata it is given (plus whatever else can be transmitted to it by virtue of the rules scope) andcan return to its calling routine at most one value, the return value, in a return statementof the form:

return returnValue;

This is unsatisfactory and restrictive if we need to return more than one value. An alternateway of calling functions, a way that can return more than one thing is called call by reference.This discussion must start, however, with a discussion of the addresses of variables, that is,where they are stored in a computer’s (virtual) memory.

6.4.1 The address of a variable

We introduce the unary operator called the address operator, &. To see its usage, considerthe following program :

//File: addressLocal.cpp

#include <iostream>


int main(void)

{ float x = 1.0f; // Declare and initialize a float

double y = 3.0; // Declare and initialize a double

int z = 2; // Declare and initialize an int

cout <<

"x’s value: " << x << " x’s address: " << &x << "\n"

"y’s value: " << y << " y’s address: " << &y << "\n"

"z’s value: " << z << " z’s address: " << &z << "\n";

return 0;

}

New stuff:

• The address operator & applied to a variable, for example, &y should be read as “theaddress of the double y”.

6.4. CALL BY VALUE, CALL BY REFERENCE, REFERENCE PARAMETERS 139

Compiling and running it results in the following output:

x’s value: 1 x’s address: 0xbffff734

y’s value: 3 y’s address: 0xbffff72c

z’s value: 2 z’s address: 0xbffff728

• When cout prints an address, it prints it out as 0x... which is the hexadecimal (hex)representation. The leading 0x is there just to tell you that what follows is in hex.

• Note the the address “stack” grows from a large address towards smaller addresses.

Note the the operating system (a Linux installation in this case) puts the first variable itencounters (in this case a 32-bit float) at address 0xbffff734. (This address, once con-verted to an integer number, is something in excess of 3 billion.) This is the address in bytes,one byte being 8 bits. A 32-bit machine has a virtual address space of 232, or 4,294,967,296bytes, often abbreviated as 4GB. Above this address, starting at 0xbffff738 in this case(This boundary is machine and O/S dependent) is where the “program environment” re-sides (we’ll discuss that later) and above that still is where the operating system resides (thekernel) occupying the upper reaches of address space up to the 4GB limit.

The next variable, a double is 64-bits or 8 bytes in length. Note that its address is smaller(by 8 bytes). The address space that is provided for local variables is called the “stack frame”and it grows downwards. Note where the next variable is located.

Each function creates its own stack frame for putting its local variables into memory, releasingit when the function terminates. This is what is meant by the stack getting “destroyed” whena function terminates. The next function call will use this space for its own stack frame.

The executable program, the “load module” is at the bottom of the address space, starting atbyte 0. This area is called the “text segment”. Above the text segment are stored initializedand uninitialized global variables. Above this is the “heap space”, the address space wheredynamic memory (memory allocated by the program) is located. It grows upwards. If yourprograms and data are so big that your stack frames and heap run into each other, youeither have to redesign your program to use less space or purchase a more expensive 64-bitcomputer, one with 264 bytes of array space. (In a few years, most computers will be 64-bitcomputers.)

Here another example program that demonstrates that global variables are located at adifferent place in memory:

//File: addressGlobal.cpp

#include <iostream>



double y = 3.0; // Declare and initialize a global double

int z = 2; // Declare and initialize a global int

int main(void)

{ float x = 1.0f; // Declare and initialize a float

cout <<

"x’s value: " << x << " x’s address: " << &x << "\n"

"y’s value: " << y << " y’s address: " << &y << "\n"

"z’s value: " << z << " z’s address: " << &z << "\n";

return 0;

}

Compiling and running it results in the following output:

x’s value: 1 x’s address: 0xbffff734

y’s value: 3 y’s address: 0x8049a70

z’s value: 2 z’s address: 0x8049a78

Note how the global variables, y and z in this case, are located starting at a lower address0x8049a70 (something like 1 billion). Note how the global stack grows up. Note also howthe double, y, occupies 8 bytes.

6.4.2 Call-by-value vs. call-by-reference

Imagine for a moment that the pow() does not exist and we wish to write a program thatcomputes the square and cube of an integer. Moreover, we wish to write a single functionthat returns both the square and cube. Using what we have learned so far, we are frustratedto discover that we can not do it! Instead we have to separate the task into two functions.Our code would look something like:

//File: returnTwice.cpp

#include <iostream>


int square(int j)

{ cout << "In function square(), j is located at " << &j << "\n";

return j * j;

}


int cube(int k)

{ cout << "In function cube(), k is located at " << &k << "\n";

return k * k * k;

}

int main(void)

{ int i = 3, iSquared, iCubed;

cout

<< "\n\nIn main():\n"

<< " i is located at " << &i << "\n"

<< "iSquared is located at " << &iSquared << "\n"

<< " iCubed is located at " << &iCubed << "\n\n"

<< "Before the call to square and cube:\n"

<< " i = " << i << "\n"

<< "iSquared = " << iSquared << "\n"

<< " iCubed = " << iCubed << "\n\n";

iSquared = square(i);

iCubed = cube(i);

cout

<< "\n After the call to square and cube:\n"

<< " i = " << i << "\n"


<< " iCubed = " << iCubed << "\n\n";

return 0;

}

where I have put in a lot of cout statements to let me know where the variables are stored.Compiling and running it results in the following output:

In main():

i is located at 0xbffff734

iSquared is located at 0xbffff730

iCubed is located at 0xbffff72c

Before the call to square and cube:

i = 3

iSquared = 134520020

iCubed = 134514715

In function square(), j is located at 0xbffff728


In function cube(), k is located at 0xbffff728

After the call to square and cube:

i = 3

iSquared = 9

iCubed = 27

Note that main()’s local variables i, iSquared, iCubed reside in main()’s stack frame.Also note that iSquared and iCubed are not initialized by main() and so the memorylocations of these variables contain nonsense bits, junk left over from the previous use ofthose memory locations, whatever process that happened to be. The first function calledis square(). It has one local variable called j which it locates in its own stack frame,starting at address 0xbffff728. square() does its work happily and returns the result ofits computation to main() which then called the function cube(). cube() has one localvariable called k which it locates in its own stack frame, starting at address 0xbffff728,the same address used by square()! Once square() returned to main(), its stack framewas destroyed and cube() did the most efficient thing, locating its stack frame at the nextavailable location, the location just vacated by square().

We now introduce the method of call-by-reference. This is best done by example, so, hereis the above program converted to a single function call that has both the usual valueparameters and also something new—reference parameters in the function’s parameter list.

//File: returnTwoThings.cpp

#include <iostream>


void squareCube(int j, int& j2, int& j3)

{ cout << "In function squareCube():\n"

<< "j is located at " << &j << "\n"

<< "j2 is located at " << &j2 << "\n"

<< "j3 is located at " << &j3 << "\n";

j2 = j * j;

j3 = j2 * j;

return;

}

int main(void)

{ int i = 3, iSquared, iCubed;

cout

<< "\n\nIn main():\n"

<< " i is located at " << &i << "\n"


<< "iSquared is located at " << &iSquared << "\n"

<< " iCubed is located at " << &iCubed << "\n\n"

<< "Before the call to squareCube:\n"

<< " i = " << i << "\n"


<< " iCubed = " << iCubed << "\n\n";

squareCube(i, iSquared, iCubed);

cout

<< "\nAfter the call to squareCube:\n"

<< " i = " << i << "\n"


<< " iCubed = " << iCubed << "\n\n";

return 0;

}

In the above example, in the parameter definition list (int j, int& j2, int& j3), thevariable j is a value parameter receiving the value of the first parameter in the call statement,the i in squareCube(i, iSquared, iCubed). In squareCube(), the variable j will beplaced in squareCube()’s stack frame. The other two parameters in the parameter definitionlist, int& j2 and int& j3, are reference parameters. What this means is that the int

variables called j2 and j3 within the function squareCube() refer to the variables locatedat the addresses of the variables given to it in the function call. When the compiler callsthe function squareCube() and sees that the parameter list in the function definition isexpecting addresses, it says, ”Aha, squareCube() is expecting addresses for iSquared andiCubed” and so the addresses are transferred to squareCube(). squareCube() puts theseaddresses in its stack frame. When the code inside squareCube() refers to the variables j2and j3, all it is really doing is renaming temporarily the memory locations referred to inmain() as iSquared and iCubed.

Compiling and running the example results in the following output:

In main():

i is located at 0xbffff734

iSquared is located at 0xbffff730

iCubed is located at 0xbffff72c

Before the call to squareCube:

i = 3

iSquared = 134519980

iCubed = 134514715


In function squareCube():

j is located at 0xbffff720

j2 is located at 0xbffff730

j3 is located at 0xbffff72c

After the call to squareCube:

i = 3

iSquared = 9

iCubed = 27

Note the memory locations of the variables local to main() and the fact that squareCube()thinks that j2 and j3 live in the same place. Note also that in squareCube(), the localvariable j is located at 0xbffff720, 12 bytes below the end of main()’s stack frame. Yet,j, an integer is only 4 bytes long. Can you guess what use squareCube() is making of those8 bytes of memory?

Finally, note that value parameters and reference parameters can be mixed in any order in theparameter list. Just make sure, in the function definition, that all the reference parametersnames are preceded by the ampersand, & symbol.

About the only real-life example that I can come up with that relates to this is my relationshipwith my stockbroker. I am the main() routine and he, the incompetent bum, is a functioncalled stockBroker(). Here is his function definition. I’ll let you figure out how I’m doingon the stock market.

//File: stockBrokerFunction.cpp

void stockBroker(double myFee, double& alexMoney)

{ double transactionCost = 200.;

while (0 < alexMoney && 0 < myFee)

{ myFee = myFee - transactionCost; //Deduct usual transaction charge

alexMoney = alexMoney * 0.95; //Make a stupid investment

}

return;

}

However, I do want you to notice that I provide him with a fee, myFee which is his to keep.However, he is playing with my money, alexMoney, a reference parameter, the value of which,I am unhappy to report, is diminishing rapidly. It is still MY money, but I have given thestockbroker the ability to change it.

6.5. THE RULES OF SCOPE 145

6.5 The rules of scope

One of the most powerful features of C++ is its implementation of the rules of scope. Thephilosophy is that information (data) should be hidden from operational code unless it reallyneeds to know about the data. This avoids the danger of a program unit changing data thatit is not supposed to—even if it calls the variables by the same name! This is how differentpeople can develop different parts of a program and as long as each programmer stays withinhis or her assigned duties, data does not get corrupted. C++ has also introduced the conceptof namespace to keep things separate. For this course we will always use: using namespace

std; although it is possible to use a different namespace. We will only use the standardnamespace in this course.

There are several hierarchies of scope:

file scope An identifier declared outside of any function definition has file scope. Thatis, “global” variables, function definitions and function prototypes (if you use them!)placed outside of any function apply for the entire file. It is perfectly fine to definean int, say, outside of any function or main. This variable can then be used by allfunctions and main. If separate files are linked together to form a program, the otherfiles will not know about file scope identifiers outside of their own files.

function scope The only identifier that is defined for the entire function is a label (anidentifier followed by a colon “:”), e.g.ThisIsALabel:

A transfer to a label can be done with the dreaded goto or within the switch statement,neither of which will be taught in this course.

Labels are defined for the whole function and undefined outside of it.

block scope A “block” is anything within braces, {...}. For example:

{ // This is a block

}

{ // This is another separate block

}

{ // This is a "yet another block"

{ // This is still another block nested

// within "yet another" block

}

}

Note the use of optional spacing to delineate the separate blocks in the above.


We have seen many examples of “block”’s—function blocks, if/else if/else blocks,while/do while/for blocks. We can also define our own blocks by surrounding state-ments with braces, {...} to define a segment of code that has its own scope.

Nested scope is treated in a very special way. Variables defined outside a scope blockwill carry into a nested scope block unless the nested scope block declares a variablewith the identical identifier. In this case the variable from the outer block is “hidden”until the inner block terminates. We will see examples of this.

function-prototype-scope One way of defining a function prototype is to include variableidentifiers. For example:

int myFunction(float a, float b, float c);

is a function prototype that names three identifiers a, b, and c. These identifiers areonly known within the function prototype. Often programmers name the variables theyare expecting in a function prototype to remember which variables go in which order.These variable names are ignored by the compiler outside of the function prototypestatement.

This is all very confusing unless we do some examples. So...let’s do some examples!

Consider the following code called scope0.cpp:

//File: scope0.cpp

#include <iostream>


void myFn(float x, float y) // MyFn declares floats x and y

{ float z; // Internal to myFn

cout << "myFn: pre-op x = " << x << ", y = " << y << "\n";

z = x; x = y; y = z; // Variable switch

cout << "myFn: post-op x = " << x << ", y = " << y << "\n";

}

int main(void)

{ float x = 1.0, y = 2.0; // Internal to main

cout << "main: pre-call x = " << x << ", y = " << y << "\n";

myFn(x, y); // Main gives values to myFn

cout << "main: post-call x = " << x << ", y = " << y << "\n";

return 0;

}


Compiling and running scope0.cpp results in the following output:

main: pre-call x = 1, y = 2

myFn: pre-op x = 1, y = 2

myFn: post-op x = 2, y = 1

main: post-call x = 1, y = 2

main declares two floats, x and y and gives myFn their values. myFn happens to call thesevariables by the same name (they could have been different). myFn’s job is to switch thevalues of these two variables. Within its own scope this has been accomplished but theswitch has not been effected in main. This is because the x and y in main and the x and y

in myFn are different.


//File: scope1.cpp

#include <iostream>


float x = 1.0, y = 2.0; // Global variables

void myFn(void) // No declarations





}

int main(void)

{ cout << "main: pre-call x = " << x << ", y = " << y << "\n";

myFn(); // Main just calls myFn


return 0;

}







In this example x and y are defined globally (outside of any function). Both main and myFn

can access them and change them. myFn’s job is to switch the values of these two variables.Within its own scope this has been accomplished and the switch has been effected in main

as well. This is because the x and y in main and the x and y in myFn are identical.


//File: scope2.cpp

#include <iostream>


float x = 1.0, y = 2.0; // Global variables

void myFn(float x, float y) // myFn declares x,y





}

int main(void)

{ cout << "main: pre-call x = " << x << ", y = " << y << "\n";

myFn(x,y); // Main transfers x and y values


return 0;

}






In this example x and y are defined globally (outside of any function). main can access themand change them. However, myFn defines its own x and y and hence the global ones arehidden and myFn cannot access them and change them.


//File: scope3.cpp

#include <iostream>



int main(void)

{ float x = 1.0, y = 2.0, z = 0.0; // Local to main

cout << " main: pre-call x, y, z = "

<< x << ", " << y << ", " << z << "\n";

{ // Start of block

float z = 42.0; // Local to {} block

cout << "block: pre-call x, y, z = "

<< x << ", " << y << ", " << z << "\n";


cout << "block: post-call x, y, z = "

<< x << ", " << y << ", " << z << "\n";

} // End of block

cout << " main: post-call x, y, z = "

<< x << ", " << y << ", " << z << "\n";

return 0;

}


main: pre-call x, y, z = 1, 2, 0

block: pre-call x, y, z = 1, 2, 42

block: post-call x, y, z = 2, 1, 1

main: post-call x, y, z = 2, 1, 0

In this example x, y, and z are defined local to main. Within main there is a nested blockthat defines its own scope. Since this block does not define x and y, these variables passinto the block. However, z is defined within the nested block and the external one is hidden.However, the nested block has changed the values of x and y when control is passed back tomain.


//File: scope4.cpp

#include <iostream>


void myFn(void)

{ float z = 0; // Internal to myFn

cout << "myFn: pre-op z = " << z << "\n";


z = 1; // Reset z

cout << "myFn: post-op z = " << z << "\n";

}

int main(void)

{ cout << "First call to myFn\n";

myFn(); // Main calls myFn

cout << "Second call to myFn\n";

myFn(); // Main calls myFn again

return 0;

}


First call to myFn

myFn: pre-op z = 0

myFn: post-op z = 1

Second call to myFn

myFn: pre-op z = 0

myFn: post-op z = 1

In this example z is defined locally local to myFn which also initialized it to zero. Every timemyFn is called, z is reset to zero and the executable statements are processed.


//File: scope5.cpp

#include <iostream>


void myFn(void)

{ static float z = 0; // Internal to myFn

cout << "myFn: pre-op z = " << z << "\n";

z = z + 1; // Add to z

cout << "myFn: post-op z = " << z << "\n";

}

int main(void)

{ cout << "First call to myFn\n";

myFn(); // Main calls myFn

cout << "Second call to myFn\n";

myFn(); // Main calls myFn again


return 0;

}


First call to myFn

myFn: pre-op z = 0

myFn: post-op z = 1

Second call to myFn

myFn: pre-op z = 1

myFn: post-op z = 2

In this example z is defined locally local to myFn which also initializes it to zero but holds itstatic. The specifier static indicates that this variable is to retain its value when it leavesa routine, preserving its value for the next entry into the routine. The initialization to zerois only effective the first time the routine is entered.

static is called a storage class and is a qualifier on the variable type that means that itsvalue is to be saved, not destroyed when the function ends execution.

Consider the following code called scope6.cpp

//File: scope6.cpp

#include <iostream>


float max(float u, float v) // function max declares u,v

{ float temp = u;

u = v;

v = temp;

cout << "max: u = " < v)

{ return u;

}

else

{ return v;

}

}

int main(void){

float x = 1.0, y = 2.0, maxxy; // Internal to main

cout << "main: x = " << x << " y = " << y << "\n";

maxxy = max(x, y);


cout << "main: max of x and y = " << maxxy << "\n";

return 0;

}


main: x = 1 y = 2

max: u = 2 v = 1

main: max of x and y = 2

main declares two floats, x and y and gives myFn their values. myFn happens to call thesevariables by different names (they could have been the same as in some of the previousexamples). myFn’s job is to switch the values of these two variables and determine whichis greater. Within its own scope this has been accomplished but the switch has not beeneffected in main. This is because the x and y in main and the u and v in myFn are different.

6.6. PROBLEMS 153

6.6 Problems

1. Another essay question

(a) In your own words numbering 50 or less, describe when you would want to makea function “call-by-reference”, rather than “call-by-value”.

(b) Then give an example of a function call-by-reference that accomplishes somethingthat you could not do using call-by-value.

(c) Give an example of

i. the function prototype,

ii. the function call, and

iii. the function definition (with an empty statement block) that you would writefor your example.

2. Simple functions

Type in the following main routine exactly as shown and add the definitions of twofunctions, cubeit0 and cubeit1, each that provides the cube of the argument x. Youmay not use the pow() math library function.

#include <iostream>

//Function cubeit0 goes here

//Function cubeit1 goes here

int main(void)

{ cout << "Input a number to be cubed: ";

double x;

cin >> x;

cout << cubeit0(x) << endl;

double result;

cubeit1(x,result);

cout << result << endl;

return 0;

}

Here is an example of how the program is intended to work:


% a.out

Input a number to be cubed: 3

27

27

%

3. Fix the bug(s)

(a) Why won’t the following code compile and/or work as intended? Fix the bug(s).

#include <iostream>


void sumInts(int a, int b)

{ int i;

i = a + b;

return i;

}

int main() {

int i = 1, j = 2;

cout << "The sum of ints i = " <

#include <cmath>


float trigs(float y, &f1, &f2)

{ f1 = sin(y); f2 = cos(y);

return 0;

}

int main(void)

{ float x = 0.0, cosx, sinx;

int returnValue = trigs(x, sinx, cosx);

cout << "sin(" << x << ") = " << sinx << ", "

<< "cos(" << x << ") = " << cosx << "\n";

6.6. PROBLEMS 155

return 0;

}

4. Predict the output

What does this code print?

#include <iostream>


void cyclic(int i, int& j, int k, int& l)

{ int m;

m = i;

i = j;

j = k;

k = l;

l = m;

}

int main(void)

{ int i = 1, j = 2, k = 3, l = 4;

cyclic(i, j, k, l);

cout << "i,j,k,l : " << i << j << k << l << "\n";

return 0;

}

5. Fix the design error(s)

(a) This code attempts to define a function that returns the maximum of two floatsand orders the two floats in descending order (bigger first, smaller second). Thiscode only works some of the time. There is one design mistake that occasionallyleads to incorrect results. Assume that the scanf statement is always executedcorrectly. You may not fix this using a library function that determines maximum.

#include <iostream>


float fMax(float f1, float f2)

{ if (f2 > f1)

{ float temp = f2; // f2 is maximum

f2 = f1; f1 = temp; // switch them

return f1;


}

else return f1; // f1 must be maximum, no need to switch

}

int main(void)

{ cout << "Input 2 float’s: ";

float a, b;

cin >> a >> b;

cout << "Maximum is: " << fMax(a,b) << "\n";

cout << "Descending order is: " << a << ", " <


void sort3(int i, int j, int k)

{ int t;

if (i > j) {t = i; i = j; j = t;}

if (j > k) {t = j; j = k; k = t;}

if (i > j) {t = i; i = j; j = t;}

}

int main(void)

{ int i = 5, j = 3, k = 1;

sort3(i,j,k);

cout << "In ascending order (smallest->largest) "

<< "the three int’s are "

<


float squareCube(float x, float xx)

{ xx = x*x;

return xx*x;

6.6. PROBLEMS 157

}

int main(void)

{ float x = 2.0, xx, xxx;

xxx = squareCube(x,xx);

cout << "Square of x: " << xx << ", Cube of x is " << xxx << "\n";

return 0;

}

6. Roots of the quadratic equation

Write a function that is compatible with the following function call:

bool twoRealRoots = twoRoots(a,b,c,root1,root2);

The function returns true if b2 − 4ac > 0 and a �= 0. Otherwise, it returns false. Ifthe function returns true, then root1 and root2 contain the roots of the quadraticequation ax2 + bx+ c = 0, which are

x1,2 =−b±

√b2 − 4ac

2a

7. Write your own√

function

The following algorithm may be used to calculate x =√a:

(a) Start with an initial guess: x = a

(b) Refine the guess: x′ = (x+ a/x)/2

(c) If x′ and x are the same, x =√a, otherwise, let x = x′ and go to step 2.

Write a function that calculates x =√a by the above method.

8. Polar coordinates

Write a function that is compatible with the following main routine:

#include <iostream>

#include <cmath>

#include <string>


//Function definition goes here

int main(void)


{ float x, y, r, t;

cout << "x, y: ";

cin >> x >> y;

cout << "x = " << x << " y = " << y << " is in the " << polar(x,y,r,t)

<< ". " << "r = " << r << " t = " << t << " degrees\n";

return 0;

}

The function, polar(), called in the second cout statement above returns a stringthat indicates

• “upper right quadrant” if x ≥ 0 and y ≥ 0

• “upper left quadrant” if x < 0 and y ≥ 0

• “lower left quadrant” if x < 0 and y < 0

• “lower right quadrant” otherwise

After the function returns, the float called r contains the radius computed asr =

√x2 + y2. Hint. r = sqrt(x*x + y*y); After the function returns, the float

called t contains the angle in degrees computed asθ = 180 tan−1(y/x)/π. Hint. Use t = 45*atan2(y,x)/atan(1.0); to calculate this.

Here is an example of how it works:

red% a.out

x, y: 1 1 //User types in "1 1" in response to the "x, y: " prompt

x = 1 y = 1 is in the upper right quadrant. r = 1.41421 t = 45 degrees

9. How many bits to an unsigned int?

The following algorithm may be used to determine the number of bits that are used torepresent an unsigned int:

(a) Starting with an unsigned int initialized to 1, keep multiplying it by 2 until itequals 0.

(b) The number of times you multiply it by 2 is related to the number of bits used torepresent the unsigned int’s.

Write a function that determines the number of bits that are used to represent anunsigned int on the computer that executes the program. Your function has to becompatible with the following main routine.

6.6. PROBLEMS 159

#include <iostream>


//Function definition goes here

int main(void)

{ cout << "The unsigned int size is " << intSize()

<< " bits on this computer\n";

return 0;

}

Here is how it works:

red% a.out

The unsigned int size is 32 bits on this computer


6.7 Projects

1. Math library functions

The following table lists some of the commonly used functions in C++’s math librarywhich can be accessed by employing #include <cmath> in the preprocessor area of aC++ file:

Math function Purpose Example usagelog(x), loge(x), ln(x) natural logarithm (base e) double x = 1.; double y = log(x);exp(x), ex exponential double x = 1.; double y = exp(x);log10(x) logarithm (base 10) double x = 1.; double y = log10(x);sin(x) sine of x double x = 1.; double y = sin(x);cos(x) cosine of x double x = 1.; double y = cos(x);√

x square root of x double x = 1.; double y = sqrt(x);|x| absolute value of x double x = 1.; double y = fabs(x);

Note that the argument to the trigonometric functions above must be in terms ofradians.360◦ (degrees) = 2π (radians).

Write a main program that generates the table on the following page similar to thatshown. In other words, you will have to have a for-loop that somehow generatesthe double’s -1.0, -0.9, ..., 1.0, calculate and output x, log(x), exp(x), log10(x),

sin(PI*x), cos(PI*x), sqrt(x), and fabs(x).

Some notes:

• Note the argument of the sin() and cos(). You were given the method togenerate a good value for π in the “Where’s Waldo?” project in the previouschapter.

• You will have to use the setw() and setprecision() functions as described inChapter 5.

• You need one other thing to allow the printing of the redundant trailing zero’s.Put in the statement

cout.setf(ios::fixed);

before your first use of cout.

• Note that the outputs -Infinity and NaN are provided automatically by cout

when the argument to the math library functions is out of range of its definition.

6.7. PROJECTS 161

x log(x) exp(x) log10(x) sin(PI*x) cos(PI*x) sqrt(x) fabs(x)

---- --------- -------- --------- --------- --------- -------- -------

-1.0 -Infinity 0.367879 -Infinity -0.000000 -1.000000 NaN 1.0





-0.5 -Infinity 0.606531 -Infinity -1.000000 0.000000 NaN 0.5





0.0 -Infinity 1.000000 -Infinity 0.000000 1.000000 0.000000 0.0

0.1 -2.302585 1.105171 -1.000000 0.309017 0.951057 0.316228 0.1

0.2 -1.609438 1.221403 -0.698970 0.587785 0.809017 0.447214 0.2

0.3 -1.203973 1.349859 -0.522879 0.809017 0.587785 0.547723 0.3

0.4 -0.916291 1.491825 -0.397940 0.951057 0.309017 0.632456 0.4

0.5 -0.693147 1.648721 -0.301030 1.000000 0.000000 0.707107 0.5

0.6 -0.510826 1.822119 -0.221849 0.951057 -0.309017 0.774597 0.6

0.7 -0.356675 2.013753 -0.154902 0.809017 -0.587785 0.836660 0.7

0.8 -0.223144 2.225541 -0.096910 0.587785 -0.809017 0.894427 0.8

0.9 -0.105361 2.459603 -0.045757 0.309017 -0.951057 0.948683 0.9

1.0 0.000000 2.718282 0.000000 0.000000 -1.000000 1.000000 1.0


2. Strive for Perfection

A “perfect” number is a positive whole number that is the sum of its “proper divisors”.The set of proper divisors of number N does not include N itself, but does include1. For example, the proper divisors of 6 are 1,2,3 and 1 + 2 + 3 = 6. So, 6 is aperfect number. In fact, it is the first perfect number. The second perfect numberis 28 = 1 + 2 + 4 + 7 + 14. The first 4 perfect numbers were known to the ancientGreek mathematicians. In fact, only 37 of them are known today! A lot is knownabout perfect numbers but much more remains a mystery. For example, the followingis known.

(a) The biggest known perfect number, the last discovered, can be written23021376 × (23021377 − 1), a number that contains 1,819,050 decimal digits.

(b) All known perfect numbers are even.

(c) All known perfect numbers are given by Euclid’s formula: N = 2k−1(2k−1) wherek is some positive integer and 2k − 1 also happens to be a prime number. This isa special kind of prime number known as a Mersenne prime. It is a really goodidea to make use of this fact in designing your computer program, describedbelow.

A lot is not known about perfect numbers. It is not known whether or not

(a) there is an infinite number of perfect numbers.

(b) all perfect numbers are even.

(c) all perfect numbers are given by Euclid’s formula.

Your job is to write a program that will:

(a) find the first 5 perfect numbers and print out their proper divisors.

(b) be composed of at least 2 functions additional to main, one that returns true ifits integer argument is a perfect number and a second function that prints outthe proper divisors of a perfect number. You may define more than 2 additionalfunctions if you wish.

(c) complete execution in less than one minute of CPU time on a run-of-the-mill,modern computer.

(d) have output that looks similar to the following:

% a.out

6 is a perfect number

6 = 1 + 2 + 3

6.7. PROJECTS 163

28 is a perfect number

28 = 1 + 2 + 4 + 7 + 14

.

.

.


3. Random integration

(a) You’ll need this later!

Write a function that accepts 3 double’s (x, x0 and x1 below) and returns thedouble (f(x, x0, x1) below) for the function:

f(x, x0, x1) =x sin(x)

1 + x∀ x0 ≤ x ≤ x1

If x is outside of the range of definition, a Nan must be returned.

Hint 0: A NaN is IEEE’s way of saying ”Not a Number”. NaN’s can be producedwhen the result of calculation using one of the floating-point representations isundefined. For example, the coding:

double zero = 0; cout << zero/zero;

produces a NaN. Try it!

Hint 1: Here’s a main routine you can use to test your function:

#include <iostream>

#include <cmath>

// The function should go here...

int main(void)

{ cout << "Input x, x0, x1: ";

double x, x0, x1;

cin >> x >> x0 >> x1;

cout << "Evaluating the function at x = " << x

<< " between the limits x0 = " << x0

<< " and " << x1 << "\n";

cout << "f(" << x << ") = " << myFunction(x, x0, x1) << "\n";

return 0;

}

(b) A random way to find an area

i. Write a function that will find the position, xmax, which is the position wherethe maximum of a function f(x) occurs. This maximum lies between twolimits x0 and x1 where x1 > x0. Your value for xmax has to be within atolerance of 10−12/|x1 − x0|. You can assume that the function has only onelocal maximum between x0 and x1, that the function f(x) ≥ 0 between x0and x1 and that the function is not a constant.

6.7. PROJECTS 165

Hint 0: Use double’s, not float’s to work on your solution.Hint 1: Life is made easy because there is only one local maximum betweenx0 and x1. That means that you can start with any x such that x0 ≤ x ≤ x1and move a little in one direction, say, x − δ, where δ is some small positivenumber. If the function is smaller there, that is, f(x − δ) < f(x), you aretesting for the maximum in the wrong direction, so try x+ δ.Hint 2: Life is made difficult because the tolerance is so small. It is notfeasible to set your initial delta to 10−12/|x1 − x0|. You will have to startwith something much larger, say, 10−1/|x1 − x0|. Once you have found theposition of the maximum within this tolerance, lower it to 10−2/|x1−x0| andso on. This incremental refinement technique is reasonably efficient. This isonly a suggestion. You can probably think of a much more efficient way todo it.

ii. Write another function that determines the area under f(x) between the twolimits x0 and x1 using a technique called random sampling. Random samplingwill be discussed in class. You will need to know the position of the functionmaximum before you can determine the area under the function.

iii. Complete the problem by using the following function:

f(x) =x sin(x)

1 + x

for 0 ≤ x ≤ π, most of which was coded in the last problem. Use 100,000samples to estimate the area.

HOW TO WORK ON THIS PROBLEM:

i. Try a few different examples for f(x) for which you know the answer (withoutknowing calculus!). Good candidates would be:

A. f(x) = x, for 0 ≤ x ≤ 1.

B. f(x) = 1− x, for 0 ≤ x ≤ 1.

C. f(x) = x, for 0 ≤ x ≤ 1/2, f(x) = 1− x, for 1/2 ≤ x ≤ 1.

ii. Once you are confident that your algorithm works, then try the assignedfunction:

f(x) =x sin(x)

1 + x

for 0 ≤ x ≤ π

iii. Write out a pseudocode to help you design your program.


Chapter 7

More Variable Types, DataAbstraction

7.1 Representation of floating-point numbers

We have seen that int’s are represented by 32 bit “words” and that they can only representa finite range of integral numbers. A signed integer covers the range −231 ≤ i ≤ (231 − 1).Floating numbers have their own restrictions. One bit is used for the sign of the number, 23bits represent the mantissa1 (the precision or granularity of the number) and 8 bits are usedto define the exponent. Consequently, float’s have a range of about 10−45 to about 10+38

and a resolution of about 1 part in 107. (The particular floating-point representation can bedifferent on different operating systems.)

double’s are represented by 64 bits. 52 for the mantissa, one for the sign and 11 for theexponent. The granularity is about 2 parts in 1016 and ranges from about 10−324 to about10+308. (The particular floating-point representation can be different on different operatingsystems.)

Some CPU manufacturers, for example Intel (which provides the CPU’s for your laboratorycomputers and, if you own them, your home computer or laptop), have adopted an IEEEstandard that represents floating point numbers with 80 bits, doing all the internal calcu-lations at precisions higher than both float’s and double’s. This variable type is calledthe long double. 64 bits are used for the mantissa, one for the sign and 15 for the ex-ponent. The granularity is about 1 part in 1019 and ranges from about 10−4965 to about10+4932. Other manufacturers have larger representations. For example, the Sun machinesmaintained by CAEN have 128-bit long double’s.

1Actually, it is assumed that the leading bit is always 1, so the representation uses 23 physical bits andone “hidden” or “virtual” bit. All the floating-point numbers have a hidden bit.

167

168 CHAPTER 7. MORE VARIABLE TYPES, DATA ABSTRACTION

This is a representation of the difference between float and double:

NENTEXPO-

8 BITS

EXPONENT11 BITS

1 BITSIGN

1 BITSIGN

MATISSA 52 BITS

DOUBLE 64-BITS

FLOAT 32-BITS

MATISSA 23 BITS

Consider the following code called precisionFloat.cpp:

//File: precisionFloat.cpp

#include <iostream>


float precision(void)

{ float one = 1.0f, e = 1.0f, onePlus;

int counter = 0;

do

{ counter = counter + 1;

e = e/2.0f;

onePlus = one + e;

}while(onePlus != one);

cout << "Converged after " << counter << " iterations\n";

return e;

}

int main(void)

{ cout << "Float resolution = " << precision() << "\n";

return 0;

}

7.1. REPRESENTATION OF FLOATING-POINT NUMBERS 169

New stuff:

1. The statement: float one = 1.0f contains the qualifier "f". This means that 1.0 isa float, typically a 32 bit constant. If the "f" were omitted, the constant would be adouble, typically a 64 bit constant.

Compiling and running precisionFloat.cpp results in the following output:

Converged after 24 iterations

Float resolution = 5.96046e-08

Consider the following code called precisionDouble.cpp:

//File: precisionDouble.cpp

#include <iostream>


double precision(void)

{ double one = 1.0, e = 1.0, onePlus;

int counter = 0;

do


e = e/2.0;

onePlus = one + e;



return e;

}

int main(void)

{ cout << "Double resolution = " << precision() << "\n";

return 0;

}

New stuff:

1. The statement

double one = 1.0, e = 1.0;

declares and initializes two double’s.


Compiling and running precisionDouble.cpp results in the following output:


Double resolution = 1.11022e-16

Consider the following code called precisionLongDouble.cpp:

//File: precisionLongDouble.cpp

#include <iostream>


long double precision(void)

{ long double one = 1.0L, e = 1.0L, onePlus;

int counter = 0;

do


e = e/2.0L;

onePlus = one + e;



return e;

}

int main(void)

{ cout << "Long double resolution = " << precision() << "\n";

return 0;

}

New stuff:

1. The statement

long double one = 1.0L, e = 1.0L;

declares and initializes two long double’s. The qualifiers "L" specify that the constant1.0 is to be considered a long double constant, an 80-bit number on my computer.

Compiling and running precisionLongDouble.cpp results in the following output:


Long double resolution = 5.42101e-20


Note that what is called Float resolution, Double resolution, or Long double resolution

in the above example outputs from the programs is the number that can NOT be resolvedwith respect to 1.

To test the range of a float consider the following code called rangeFloat.cpp:

//File: rangeFloat.cpp

#include <iostream>

#include <iomanip>


int main(void)

{ float x = 1.0f, x0;

float y = 1.0f, y0;

int counter = 0;

do


x0 = x; x = x*2.0f; // Multiply by 2

y0 = y; y = y/2.0f; // Divide by 2

cout << setw(4) << counter << ": "

<< setw(12) << x << " :: "

<< setw(12) << y << "\n";

}while(x0 != x || y0 != y);

return 0;

}

Compiling and running rangeFloat.cpp results in the following output:

1: 2 :: 0.5

2: 4 :: 0.25

3: 8 :: 0.125

4: 16 :: 0.0625

5: 32 :: 0.03125

.

.

.

125: 4.25353e+37 :: 2.35099e-38

126: 8.50706e+37 :: 1.17549e-38

127: 1.70141e+38 :: 5.87747e-39

128: Inf :: 2.93874e-39

129: Inf :: 1.46937e-39

130: Inf :: 7.34684e-40


.

.

.

147: Inf :: 5.60519e-45

148: Inf :: 2.8026e-45

149: Inf :: 1.4013e-45

150: Inf :: 0

New stuff:

1. The output Inf is the IEEE’s way of representing a number that is too large to berepresented. Inf has its own distinctive bit pattern.

2. Note that the float’s appear to have more “range” for small numbers than largenumbers. IEEE allows for a gradual decrease to zero. These numbers below about6e-39 are called “sub-normal”, with fewer bits of precision. The leading bits of themantissa are padded with zeros for sub-normal numbers.

To test the range of a double consider the following code called rangeDouble.cpp:

//File: rangeDouble.cpp

#include <iostream>

#include <iomanip>


int main(void)

{ double x = 1.0, x0;

double y = 1.0, y0;

int counter = 0;

do


x0 = x; x = x*2.0; // Multiply by 2

y0 = y; y = y/2.0; // Divide by 2


<< setw(12) << x << " :: "

<< setw(12) << y << "\n";

}while(x0 != x || y0 != y);

return 0;

}

Compiling and running rangeDouble.cpp results in the following output:


1: 2 :: 0.5

2: 4 :: 0.25

3: 8 :: 0.125

4: 16 :: 0.0625

5: 32 :: 0.03125

.

.

.

1021: 2.24712e+307 :: 4.45015e-308

1022: 4.49423e+307 :: 2.22507e-308

1023: 8.98847e+307 :: 1.11254e-308

1024: Inf :: 5.56268e-309

1025: Inf :: 2.78134e-309

.

.

.

1073: Inf :: 9.88131e-324

1074: Inf :: 4.94066e-324

1075: Inf :: 0

To test the range of a long double (at least an IEEE 80-bit one as implemented on mymachine) consider the following code called rangeLongDouble.cpp:

//File: rangeLongDouble.cpp

#include <iostream>

#include <iomanip>


int main(void)

{ long double x = 1.0L, x0;

long double y = 1.0L, y0;

int counter = 0;

do


x0 = x; x = x*2.0L; // Multiply by 2

y0 = y; y = y/2.0L; // Divide by 2


<< setw(12) << x << " :: "

<< setw(12) << y << "\n";

}while(x0 != x || y0 != y);

return 0;


}

Compiling and running rangeLongDouble.cpp results in the following output:

1: 2 :: 0.5

2: 4 :: 0.25

3: 8 :: 0.125

4: 16 :: 0.0625

5: 32 :: 0.03125

.

.

.

16382: 2.97433e+4931 :: 3.3621e-4932

16383: 5.94866e+4931 :: 1.68105e-4932

16384: Inf :: 8.40526e-4933

16385: Inf :: 4.20263e-4933

.

.

.

16443: Inf :: 1.45808e-4950

16444: Inf :: 7.2904e-4951

16445: Inf :: 3.6452e-4951

16446: Inf :: 0

Now that’s a lot of numbers. Try these example codes on your own machine. If you getoutput that is different, particularly for the long double’s, I’d like to know about it!

7.1.1 When is one not one? Floating point anomalies.

According to mathematics, a very strange way of writing one would be

N∑1

1

N

Is this true for computers? No!

The following code demonstrates the effect.

//File: oneNotOne.cpp

#include <iostream>

#include <iomanip>

7.2. CHAR: THE CHARACTER VARIABLE 175


int main(void)

{ cout << "N? ";

double N;

cin >> N;

float fraction = 1.0/N;

cout << "Will compute " << fraction << " * " << N << " = ";

float sum = 0;

for (int i = 1; i <= N; i = i + 1) sum = sum + fraction;

cout << setprecision(19) << sum << "\n";

return 0;

}

Try running this code for small values of N and big values of N , say N = 107 or 108. Youwill get answers near 1 for small N , but something entirely different for big N ! Why? Howcould you fix this?

Recall that the representation of floating-point numbers is inexact in computers. If you areusing float’s with a precision of about 1 part in 107, you will start to notice anomalies ifyou try to add two things together that differ in size by 6 or 7 orders of magnitude. Theseanomalies can be readily observed with float’s. Converting to double’s or long double’streats the symptom but does not cure the disease! It is always there but often not noticeablewith higher precision numbers. So why not use double’s or long double’s all the time?Higher precision costs in terms of storage. Sometimes memory requirements dictate theuse of lower precision. There are speed/performance issues to consider as well. This is yetanother one of the compromises one has to make in designing computer programs.

7.2 char: the character variable

• A single character can be defined as the character type: char

• char is a data type used to encode single characters

• The following line declares a character called ch

char ch;

• The following assigns the character ’a’ to the char variable. A literal character is acharacter between single ’ quotes.


ch = ’a’;

• The following line declares a character called ch and assigns it the character ’a’

char ch = ’a’;

• The following statement contains an error. What is the error?

ch = "a";

• Some special characters:

– ’\n’ is a “newline”

– ’\t’ is a “horizontal tab”

– ’\0’ is a “null”. It means nothing! We have to make a special character thatmeans nothing!

– There are lots of others.

• The inner secret of char’s? They are really integers!But, they are really integers but only one byte (8-bits) long.Thus there can be at most 28 = 256 of them.

• The following are all valid statements:

char c;

int i;

i = 65;

c = i;

c = 99;

• Consider the program :

//File: chars.cpp

#include <iostream>


int main(void)

{ for(int i = 0; i < 256 ; i = i + 1)

{ char c = i;

if (i <= 127)

7.2. CHAR: THE CHARACTER VARIABLE 177

cout << "Int " <" << c << " in ASCII\n";

else

cout << "Int " <" << c << " beyond ASCII\n";

}

return 0;

}

Try running this program and see what it gives you!

• The characters corresponding to integers 0—127 are the ASCII standard characterset. (ASCII = American Standard Code for Information Interchange) which is univer-sally adopted in English-speaking nations. The printing characters are represented byASCII codes 32–126. Some of the ASCII codes do special things or represent “special”characters. For example, 7 is the “alert” or “bell”, 8 is the “backspace”, 9 is the “hor-izontal tab”, 10 is a “new line”, 11 is a “vertical tab”, 13 is a “carriage return”, 127 isthe “delete”. Note the effect of some of these when running chars.cpp.

• The extended ASCII character set corresponding to integers 0—255 is internation-ally adopted and permits many of the characters from some foreign languages to berepresented.

• Movement is afoot to accept the UNICODE standard which codes characters in 2bytes (16 bits) with 216 = 65536 possibilities. This is enough to code not only everylanguage that exists, but even every language that has ever existed. Even Egyptianhieroglyphics has a UNICODE representation. There is probably no need to extendbeyond 16-bit representation until we start making contact with extra-terrestrial lifeforms, well, at least ones that want to communicate with us for whatever reason.

7.2.1 Character strings; the string class

• The string class requires the use of #include <string>

• You have seen them already!cout << "This is a string!\n";

• Declaring a string called str and assigning it a value:string str = "This is a string!\n";

• A character string is really kind of a vector in disguise.str[0] is "T"str[1] is "h"etc.


• Strings can be read in with cin and printed with cout. The cin operation continueson the string until it sees a space or a new-line character (hitting the Return key).Here is an example that will be discussed in class:

//File: string1.cpp

#include <iostream>

#include <string>


int main(void)

{ string response;

do

{ cin >> response;

cout << response + "\n";

}while (’.’ != response[0]); //Note vector notation

return 0;

}

It is important to understand that the space character is considered by the cin functionto signal the end of a string.

• The fact that a space character is considered by the cin operation to signal the endof a string can sometimes be a nuisance if you want to input spaces. For example, youmight want an entire sentence to be considered a single string. The getline() stringclass member function (as declared by <string>) can be used for this purpose. Hereis an example that will be discussed in class:

//File: string2.cpp

#include <iostream>

#include <string>


int main(void)

{ string sentence;

do

{ getline(cin,sentence);

cout << sentence + "\n";

}while (’.’ != sentence[0]);

return 0;

7.3. THE VECTOR CLASS 179

• The length() member function returns the length of a string, the number of charactersthat it contains. The substr() member function extracts a sub-strings from a string.Note that strings can be concatenated with the addition operator, although at leastone of the operands must be a string variable. Here is an example that demonstratesthese three new things.

//File: string3.cpp

#include <iostream>

#include <string>


int main(void)

{ cout << "Type in a string: ";

string r1;

cin >> r1;

cout << "Your response was " << r1.length() << " chars long\n";

cout << "Type in another string: ";

string r2;

cin >> r2;

cout << "Your response was " << r2.length() << " chars long\n";

cout << "Putting them together: " << r1 + " " + r2 << "\n";

cout << "Slicing and splicing them up: "

<< r1.substr(0,2) + r2.substr(4,7) << "\n";

return 0;

}

7.3 The vector class

7.3.1 Introduction to vectors

So far we have been dealing with individual float’s and int’s, declaring them individually,naming them individually, manipulating them one at a time, individually. Individuality isnice, but sometimes it is more efficient to refer to a group of similar individuals in a collectivemanner. In many applications we wish to use data, many of the same type, in some sort ofsystematic or regular way.


For example, a function of x is called f(x). An operation on f(x), such a multiplicationby a constant factor operates on all the elements of f(x), effectively an infinite number ofoperations! A more tangible example is; say a kind-hearted professor wanted to raise all thegrades in the class by 5%. If he had a collection of the grades in an ordered list, it would bea lot easier (and more likely to happen) if he could multiply the whole list at once by 1.05,rather than each grade on an individual basis.

To carry this example a little further, our class list is made up of 217 individuals (as in thecase of ENG101/200). Each of these individuals is assigned a different grade at the end of thecourse. Let’s say we wanted to calculate the class average. A computation of a class averagewould involve accessing all this data in some regular fashion using a computer program.

C++ provides a way of doing this by allowing you to define a vector—a symbol that isassociated with many objects.

7.3.2 Declaring and using vectors

Suppose I wanted to declare variables associated with the individual grades in this class. Icould do the following:

int gradeForStudent1 = 98;



.

.

.


This is a really dumb, repetitive way to do it! Anything that seems really repetitive fora human can probably be done in a better way by a machine. C++ allows you to defineeverything at once using the following syntax:

vector<int> gradeForStudent(217);

which declares, in one fell swoop, 217 int’s. (We could also have made them float’s ordouble’s. We can make vectors out of any variable type.) So, we can refer to the collectionof grades by a single identifier, gradeForStudent, in a collective fashion.

But we still have work to do. We still have to enter the numbers somehow into the computerprogram. One way is through assignment statements:

gradeForStudent[0] = 98; /* 1st student */


gradeForStudent[1] = 79; /* 2nd student */

gradeForStudent[2] = 88; /* 3rd student */

.

.

.

gradeForStudent[216] = 91; /* 217th student */

which also seems a little dumb, but indicates that we can also refer to the individuals by anumerical index. (For example, 7of9 in the Borg collective. If you are not a Trekkie, pleaseignore the last statement. However, for some strange reason, 7of9 is the only individual Borgthat I can name.)

In fact, for data of this nature, there is always some boring work to be done—typing in allthe grades. Unless there is an automatic way to generate the components of the vector, someperson has to do the boring job of typing it all in. This is called data entry.

The above example, however, shows that vector indices (the integer inside the []’s) start with zero.This leads to “off-by-one” errors that plague all novice (and some experienced) programmersin C++. So, just remember that the n’th element in a vector is referred to by the index n

- 1.

Another way of entering the data to the vector is by using an input loop. For example:

#include <iostream>

#include <vector>

.

.

.

const int NumberOfStudents = 217;

vector<int> gradeForStudent(NumberOfStudents);

for(i = 0; i < gradeForStudent.size(); i = i + 1)

{ cout << "Input grade for student number <> gradeForStudent[i];

}

gets the input done in another way, although someone still has to do all the typing.

There is some new stuff here that needs a little explanation. The vectors that we have justintroduced are really a class in C++. A class is defined as data and functions that operate onthe data within the class. The function size() is one of the vector class member functions.The syntax to have a member function operate on a data variable in its class is: name thedata variable and connect it via the period, “.”, and then name the member function withthe appropriate argument list, in this case nothing. The size() member function returnsthe number of elements in the vector, in this case, 217.


There is one other new thing in the above code snippet, the introduction of the const

qualifier to a variable. All this means is that the variable is to be held constant after itsinitialization. Any attempt to change it would result in an error. This is a safe way to keepvector sizes safe from programmer error if they are known to be fixed.

The best way to input the numbers is to input them in a separate file and read this filein during the execution of the code. This would make the code more general and the codeitself would not have to be concerned with having to enter the data faithfully. And, anotherprogram could make use of this data. We will learn how to read and write files later on inthe course.

7.3.3 Vector syntax and rules:

• You need to #include <vector> to use vectors.

• The name of a vector (e.g. gradeForStudent) follows the naming convention for iden-tifiers.

• Vectors can be “global” or “local”, just like individual variables, depending on wherethey are declared.

• The “index” also called the “subscript” is the thing that goes inside the [ ]’s.

• What goes inside the [ ]’s is an integer or an integer expression.

• If you violate the bounds of a vector (write data either below it or above it), one oftwo things can happen

– Either your program will crash, or

– The data occupying memory adjacent to the vector will be corrupted, the programwill carry on and operate on or with corrupted data.

At least in the first situation you know something is wrong. In the second you don’tand that could be dangerous.

• The i’th element of vector someVector is someVector[i - 1]

• someVector[someIntegerExpression] denotes a value. It can go on the left of anassignment operator “=”

• When a vector is declared all of its elements are set to zero.

• Once declared, it is best to think of vectors as regular variables. For example,

vector<float> vector2 = vector1;


where vector1 was previously defined, creates a new vector called vector2 with thesame size as vector1 and copies all the elements of vector1 to vector2.

Example: Computing averages

Suppose we have a class of 10 students. I want to compute the average grade of a class test.Here is a way to do it.

//File: sumGradesGood.cpp

#include <iostream>

#include <vector> //Need this to use the vector "class"


int main(void)

{ const int MaxNoStudents = 10;

vector<int> grade(MaxNoStudents);

int sum = 0;

for (int i = 0; i < grade.size(); i = i + 1)

{ cout << "Grade for student #" <> grade[i];

sum = sum + grade[i];

}

cout << "\nGrade report:\n\n";


cout << "Grade for student #" <(sum)/grade.size() << "\n";

return 0;

}

And here is a sample use of the code:

Grade for student #1: 1











Grade report:











Class avg is : 5.5

Example: A more refined way of computing averages

The above example requires that the C++ code know in advance how many students thereare in the class. So, if the number of students change, the code has to be recompiled. Whynot ask the user of the program to input the number of students, then declare the vectorand then do the calculation? Why not? Here is a better way to do it.

//File: sumGradesBetter.cpp

#include <iostream>



int main(void)

{ cout << "Number of students writing the test: ";

int nStudents;

cin >> nStudents;

vector<int> grade(nStudents);

int sum = 0;



{ cout << "Grade for student #" <> grade[i];

sum = sum + grade[i];

}



cout << "Grade for student #" <(sum)/grade.size() << "\n";

return 0;

}

A example of its use is:

Number of students writing the test: 3




Grade report:




Class avg is : 7.66667

Example: An even better way

I’m still dissatisfied with the above program. It requires me to count how many studentswrote the test. Why not have the vector simply grow in size as I enter more data? Why not?Here is an even better way to do it. In this case, the input loop terminates with a negativegrade or a grade larger than 100.

//File: sumGradesBest.cpp

#include <iostream>




int main(void)

{ vector<int> grade;

double sum = 0, score;

do

{ cout << "Input a student’s grade: ";

cin >> score;

if (0. <= score && 100. >= score)

{ sum = sum + score;

grade.push_back(score);

}

}while(0. <= score && 100. >= score);



cout << "Grade for student #" << i + 1 << ": \t" << grade[i]

<< "\n";

cout << "Class avg is : " << sum/grade.size() << "\n";

return 0;

}

An example of its use is:

Input a student’s grade: 79




Input a student’s grade: -1

Grade report:





Class avg is : 73.75

The new features of the above program is a new member function of the vector class, seenin the statements:


.

.

.

vector<int> grade;

.

.

.

grade.push_back(score);

.

.

.

Note that the first statement above simply declares a vector without a size! We are free to dothis. All it does is bind the identifier grade to a vector variable type. The member functionpush back(score) increases the size of the vector by one and initializes the new locationwith its argument, in this case, the value of score. There is another member function calledpop back(), with no argument, that reduces a vector’s size by one.

7.3.4 Vectors and functions

Remember that it is usually useful to think of vectors as simple variables. This is especiallytrue for vectors and functions. Take all the rules you have learned about functions andvariables and apply them to functions and vectors. To demonstrate this, consider the fol-lowing example that demonstrates numerical integration. All the algorithm does is integratea simple function in three different ways. First, it puts the function into a vector with acertain number of points based on an evenly divided mesh, or grid. Then it forms rectanglesin 3 different ways, using for its height the left mesh point, the right mesh point and themidpoint. Can you guess why the midpoint method is so much better?

//File: integrate.cpp

#include <iostream>

#include <cmath>

#include <vector>


double integrateLeft(vector<double> f, vector<double> x)

{ double sum = 0;

for (int i = 0; i < f.size() - 1; i = i + 1)

sum = sum + f[i]*(x[i + 1] - x[i]);

return sum;


}

double integrateRight(vector<double> f, vector<double> x)

{ double sum = 0;


sum = sum + f[i + 1]*(x[i + 1] - x[i]);

return sum;

}

double integrateMiddle(vector<double> f, vector<double> x)

{ double sum = 0;


sum = sum + 0.5*(f[i] + f[i + 1])*(x[i + 1] - x[i]);

return sum;

}

int main(void)

{ cout << "Number of points in the integration mesh: ";

int mesh;

cin >> mesh;

vector<double> f(mesh), x(mesh);

for(int i = 0; i < mesh; i = i + 1)

{ x[i] = static_cast<double>(i)/(mesh - 1);

f[i] = exp(x[i]);

}

double integral = integrateLeft(f, x);

cout <<

"left = \t\t" << integral << " cf " << exp(1.0) - 1.0

<< " diff = " << integral - (exp(1.0) - 1.0) << "\n";

integral = integrateRight(f, x);

cout <<

"right = \t" << integral << " cf " << exp(1.0) - 1.0


integral = integrateMiddle(f, x);

cout <<

"midpoint = \t" << integral << " cf " << exp(1.0) - 1.0


return 0;

7.4. PROBLEMS 189

}

7.4 Problems

1. The inner product

Complete the program started below. You must complete the function innerProduct()which computes the inner product of two vectors, x and y. The inner product is thesum x0y0 + x1y1 + ...+ xn−1yn−1 where xi is the i+ 1’th element of x and the numberof elements in the vector is n. Here is a function stub and a main routine that youmust use to test your function:

#include <iostream>

#include <vector>

#include <cstdlib>


float innerProduct(const vector<float>& a, const vector<float>& b)

{ //Why did the function use the const qualifiers in the parameter list

//and pass the vectors by reference?

//Answer the question and insert your code after this line.

}

int main(void)

{ int vectorLength = 10;

vector<float> x (vectorLength);

cout << " x: ";

for (int i = 0; i < vectorLength; i = i + 1)

{ x[i] = rand()%100;

cout << "\t" << x[i];

}

cout << "\n";

vector<float> y(vectorLength);

cout << " y: ";


{ y[i] = rand()%10;

cout << "\t" << y[i];

}


cout << "\n";

cout << "Inner product is: " << innerProduct(x,y) << "\n";

return 0;

}

In the comments of your function, answer the question: Why did the function use theconst qualifiers in the parameter list and pass the vectors by reference?

2. Maximum and minimum

Complete the program started below. You must complete the function minMax() whichreturns both the minimum and the maximum elements of the vector. Here is a functionstub and a main routine that you must use to test your function:

#include <iostream>

#include <vector>

#include <cstdlib>


void minMax(const vector<int>& c, int& min, int& max)

{ //Why were min and max given as "reference parameters"?


}

int main(void)


vector<int> a(vectorLength);

cout << " a: ";


{ a[i] = rand();

cout << "\t" << a[i];

}

cout << "\n";

int minimum, maximum;

minMax(a, minimum, maximum);

cout << "The minimum array element is: " << minimum << "\n";

cout << "The maximum array element is: " << maximum << "\n";

7.4. PROBLEMS 191

return 0;

}

In the comments of your function, answer the question: “Why were min and max givenas ‘reference parameters’?”

3. Nothing negative

Complete the program started below. You must complete the function nothingNegative()which accepts a vector called bothSigns containing positive and negative integers andreturns a vector of equal or lesser length that contains only the non-negative integersof the vector bothSigns. Here is a function stub and a main routine that you mustuse to test your function:

#include <iostream>

#include <vector>

#include <cstdlib>


vector<int> nothingNegative(const vector<int>& a)

{ //Why can this program, when compiled with just the function stub,

//go into what looks like an infinite loop?


}

int main(void)


vector<int> bothSigns(vectorLength);

cout << " Input vector: ";


{ bothSigns[i] = rand()%201 - 100;

cout << "\t" << bothSigns[i];

}

cout << "\n";

vector<int> noNegs = nothingNegative(bothSigns);

cout << " Output vector: ";

for (int i = 0; i < noNegs.size(); i = i + 1)

{ cout << "\t" << noNegs[i];


}

cout << "\n";

return 0;

}

In the comments of your function, answer the question: “Why can this program, whencompiled with just the function stub, go into what looks like an infinite loop?”

4. Vector manipulations

Write a function that accepts one called-by-reference vector of int’s and returns anothervector of int’s. Upon entry to the function, the vector in the parameter list can beany size and contains int’s that can have any value. Upon return, the vector in theparameter list contains only the positive int’s (> 0) in the same order. Upon return,the returned vector contains only the negative int’s (< 0) in the same order as theyappeared in the original called-by-reference vector in the parameter list, when thefunction was entered.


Vector in the parameter list supplied to the function:

0 -6 -8 3 4 -8 0 8 -9 -8 8

That vector after the function returns:

3 4 8 8

Returned vector after the function returns:

-6 -8 -8 -9 -8

5. Push and pop

This is a more challenging version of the previous question. Complete the programstarted below. Write a void function that accepts two called-by-reference vectors ofint’s. Upon entry to the function, the first vector can be any size and contains int’sthat can have any value. The second vector is empty, its size is 0. Upon return,the first vector contains only the positive int’s (> 0) in the same order. Upon re-turn, the second vector contains only the negative int’s (< 0) in the same order asthey appeared in the original first vector, when the function was entered. Important:Do this without declaring any new vectors.

#include <iostream>

#include <vector>

#include <cstdlib>

7.4. PROBLEMS 193

#include <ctime>


//Function definition goes here...

int main(void)


int size = rand()%20 + 10; //size is a RN between 10 and 29

vector<int> in(size), out;

for (int i = 0; i < in.size(); i = i + 1)

in[i] = rand()%19 - 9; //RN between -9 and +9

cout << "Original vector is size " << in.size() << endl;


cout << in[i] << " ";

cout << "\n\n";

extractNegs(in, out);

cout << "Returned vector is size " << out.size() << endl;

for (int i = 0; i < out.size(); i = i + 1)

cout << out[i] << " ";

cout << "\n\n";

cout << "Modified vector is size " << in.size() << endl;


cout << in[i] << " ";

cout << "\n\n";

return 0;

}

6. More vector manipulations

Write a function that returns a vector<int> and accepts one called-by-referencevector<int>. Upon entry to the function, the vector in the parameter list can beany size and contains int’s that are positive, negative and 0. Upon return, the vectorthat is returned contains the indices of the original vector whose original values werenegative. Upon return, the original vector in the parameter list is modified so thatit only contains its original positive (> 0) int’s in the same order that they appearedin the original vector. Write your function below or on the next page. It should be


compatible with the main routine below..

//Precompiler code omitted for brevity

//Function goes here

int main(void)


int size = rand()%11 + 10; //size is random between 10 and 20

vector<int> in(size);


in[i] = rand()%19 - 9; //vector element is random between -9 and 9

cout << "Original vector is size " << in.size() << endl;

cout << "Original vector: ";

for (int i = 0; i < in.size(); i = i + 1) cout << in[i] << " ";

cout << endl;

vector<int> negs = removeNegs(in);

cout << "Modified vector is size " << in.size() << endl;

cout << "Modified vector: ";

for (int i = 0; i < in.size(); i = i + 1) cout << in[i] << " ";

cout << endl;

cout << "Returned vector is size " << negs.size() << endl;

cout << "Returned vector: ";

for (int i = 0; i < negs.size(); i = i + 1) cout << negs[i] << " ";

cout << endl;

return 0;

}


red% a.out

Original vector is size 13

Modified vector: -9 0 4 9 9 9 0 3 3 -1 -6 5 0

Modified vector is size 7

Modified vector: 4 9 9 9 3 3 5

Returned vector is size 3

Returned vector: 0 9 10

7.4. PROBLEMS 195

7. Perfect squares

Complete the program started below. Write a function that returns a vector<int>

and accepts one constant called-by-reference vector<int>. Upon entry to the function,the vector in the parameter list can be any size and contains int’s that are positive(> 0). Upon return, the vector that is returned contains the indices of the first vectorwhose values are perfect squares.

#include <iostream>

#include <vector>

#include <cstdlib>

#include <cmath>

#include <ctime>


//Function definition goes here...

int main(void)


int size = rand()%20 + 10; //size is a RN between 10 and 29

vector<int> in(size);


{ in[i] = rand()%90 + 10;

cout << in[i] << " ";

}

cout << endl;

vector<int> out = squareIndex(in);

cout << "Returned vector is size " << out.size() << endl;

for (int i = 0; i < out.size(); i = i + 1) cout << out[i] << " ";

cout << endl;

return 0;

}

8. Sort this out

Complete the program started below. You must complete the function sort() thathas the void return type. This function accepts a vector called unsorted containingrandom int’s. Here is a function stub and a main routine that you must use to testyour function:


#include <iostream>

#include <vector>

#include <cstdlib>


void sort(vector<int>& a){}

int main(void)

{ cout << "Input the minimum and maximum random int: ";

int minRange, maxRange;

cin >> minRange >> maxRange;

cout << "Input the length of the vector: ";

int vectorLength;

cin >> vectorLength;

vector<int> unsorted(vectorLength);

cout << " Vector beforehand: \t";


{ unsorted[i] = minRange + rand()%(maxRange - minRange + 1);

cout << unsorted[i] << " ";

}

cout << "\n";

sort(unsorted);

cout << " Vector afterhand: \t";

for (int i = 0; i < unsorted.size(); i = i + 1)

{ cout << unsorted[i] << " ";

}

cout << "\n";

return 0;

}

9. A prime example

Complete the program started below. You must complete the function getPrimes()

that has the vector<int> return type. This function accepts an int called NPrimes

that is the number of consecutive prime numbers starting with 2. The return vectorcontains these prime numbers in order.

You must make your function efficient by using something similar to the following testto eliminate numbers that are not prime:

7.4. PROBLEMS 197

if (0 == n%divisor) nIsPrime = false;

where N is the candidate prime and divisor is a prime number. The divisor mustnot be non-prime.

#include <iostream>

#include <vector>

#include <cstdlib>

vector<int> getPrimes(const int NPrimes){}

int main(void)

{ cout << "Input the number of primes to find: ";

int NPrimes;

cin >> NPrimes;

vector<int> primes = getPrimes(NPrimes); //Get the primes

cout << "The first " << primes.size() << " prime numbers: ";

for (int i = 0; i < primes.size(); i = i + 1)

{ cout << primes[i] << " ";

}

cout << "\n";

return 0;

}

10. No repeats

Complete the program started below. You must complete the function noRepeats()

that has the void return type. This function accepts a vector called bothSigns con-taining random int’s and eliminates any values in it that are repeated. Here is afunction stub and a main routine that you must use to test your function:

#include <iostream>

#include <vector>

#include <cstdlib>


void noRepeats(vector<int>& a){}

int main(void)


{ cout << "Input the minimum and maximum random int: ";

int minRange, maxRange;

cin >> minRange >> maxRange;

cout << "Input the length of the vector: ";

int vectorLength;

cin >> vectorLength;

vector<int> hasRepeats(vectorLength);

cout << " Vector beforehand: \t";


{ hasRepeats[i] = minRange + rand()%(maxRange - minRange + 1);

cout << hasRepeats[i] << " ";

}

cout << "\n";

noRepeats(hasRepeats);

cout << " Vector afterhand: \t";

for (int i = 0; i < hasRepeats.size(); i = i + 1)

{ cout << hasRepeats[i] << " ";

}

cout << "\n";

return 0;

}

Chapter 8

More Data Abstraction, Arrays,Structures

8.1 Arrays

An array is an extension of the vector concept. An array can represent a list of objects witha single index, for example, Ai, for i = 1, 2, 3...N . For numerical computation, so calledone-dimensional arrays are used to represent lists of objects (e.g. a list of grades). Twoone-dimensional arrays (e.g. x[i],f[x]) can represent a function f(x) by specifying a gridof points along the x-axis, x[i], or xi, and the function evaluated at those grid points, f[i],or fi = f(xi). These are the most common uses of one-dimensional arrays.

Vectors, which we encountered in the previous lecture, are really one-dimensional arrays.However, arrays are a lower level construct that comes from C. Simply stated, arrays are or-dered collections of things you already know about, for example, int’s, float’s and double’swith a few special rules related to indexing, initializing and communicating with functions.A vector, on the other hand, is a standard class in C++ and it comes with powerful memberfunctions like size(), push back() and pop back(), that operate on the data in vectors. Forthis reason, we will not investigate one-dimensional arrays in detail in this course, choosinginstead to focus on the more powerful vector class when dealing with 1-D applications.

The multi-dimensional array is an extension of the one-dimensional array concept. A two-dimensional array can be thought of as a grid or table of values, a set of rows and columns,each with its own value. There are many expressions of these. A two-dimensional array canrepresent a checkerboard, a chessboard, the board in game of Battleship or Minesweeper, afootball field, or the surface of the earth. (Yes, it is two-dimensional, Christopher; it’s justnot flat!) We could think of many examples, there being no limit to human creativity. Afterthis lecture, you will have enough syntax tools and rules to program a game of chess. Witha little more understanding of atmospheric physics (not programming), after this lecture

199

200 CHAPTER 8. MORE DATA ABSTRACTION, ARRAYS, STRUCTURES

you could write a program that predicts the weather! After only about 20 lectures we havegotten this far! Now this is getting interesting! You have the tools and the rules, now applyyour creative thinking to design algorithms which themselves are only composed of threethings—the three pillars of algorithm design: sequence, branch and loop. It really just boilsdown to this. It really is that simple and wonderful.

In more abstract terms, a two-dimensional array can represent a table of objects with twoindices, for example, Ai,j. For numerical computation in science and engineering, two-dimensional arrays are used to represent tables of objects as in the examples above. Two-dimensional arrays (e.g. x[i][j],f[i][j]) can represent a function of two variables, f(x, y),by specifying a grid of points in the xy-plane, x[i][j], or xi,j , and the function evaluatedat those grid points, f[i][j], or fi,j = f(xi,j). However, the most interesting ones for scien-tists and engineers are “square” two-dimensional arrays, also called matrices. In this course,we will restrict our discussion to 2-dimensional arrays. There is nothing new in three andhigher dimensions that we will not see in studying 2-dimensional ones. And, the rules forone-dimensional arrays can be obtained easily from two-dimensional arrays.

Unfortunately, C++ has not defined a new class to describe these objects. So we will haveto grapple with the more primitive array inherited from C. Later on in the course, we will bestudying Matlab. Matlab is built on two-dimensional arrays as the the fundamental variabletype and has developed a lot of functionality for it, which we will discover later. Matlab isa very sophisticated program (used by professional engineers worldwide). However, if youcan develop the application in C++, it will always run faster, sometimes much faster. So,we spend some effort learning how to work with two-dimensional arrays in C++. Besides,doing so will be an excellent exercise in algorithmic thinking—the prime reason for teachingthis course.

The ANSI standard calls for arrays of at least 12 dimensions. Believe it or not, there areactually applications for this!

OK, enough with the sermon, already. Let’s dive into the syntax before we learn how toexploit it for creative expression. Some of the syntax was already introduced above.

8.1.1 Declaring a 2-dimensional array

Here’s an example of how to declare and initialize a two-dimensional array:

const int NROWS = 3;

const int NCOLUMNS = 2;

int a[NROWS][NCOLUMNS];

This is called a “3 by 2 array” or more generally as an NROWS by NCOLUMNS array. It is usefulto visualize 2-dimensional arrays with the columns numbered 0 to NCOLUMNS - 1 going from

8.1. ARRAYS 201

left to right and the rows numbered 0 to NROWS - 1 going down the page. Of course, thisvisualization is completely arbitrary but helps serve as a memory aid.

a[0][0] a[0][1]

a[1][0] a[1][1]

a[2][0] a[2][1]

Another useful way to think about it is as a one-dimensional array of other one-dimensionalarrays. Indeed, the notation a[NROWS][NCOLUMNS] is very suggestive of this! We can visualizethis as vectors of length NROWS aligned vertically and columns of length NCOLUMNS alignedhorizontally. Use whatever visual aid works for you. The above two suggestions work forme.

8.1.2 Initializing a 2-dimensional array

Initializing a 2-dimensional array in the declaration statement can be done in a variety ofways. Here are some examples:



int a[NROWS][NCOLUMNS] = {0};

Sets all of the array elements to zero. That is:

0 0

0 0

0 0

This feature is important to remember as it saves a lot of typing, a particular passion ofmine.

The brace notation may also be used:



int a[NROWS][NCOLUMNS] = {{1,2}, {3,4}, {5,6}};

The result is:


1 2

3 4

5 6

The brace notation is nice because it may be used to initialize the array a little more sug-gestively. For example,



int a[NROWS][NCOLUMNS] =

{ {1,2},

{3,4},

{5,6}

};

with the same result.

Note the use of nested braces. If a brace is not filled completely, it is filled with zeros. Forexample,



int a[NROWS][NCOLUMNS] = {{1}, {3}, {5}};

results in

1 0

3 0

5 0

Note that it is important, when initializing arrays in this fashion, to make the array sizedeclared with int’s that are fixed with the const qualifier. Otherwise the compiler willcomplain and not complete. Arrays can be declared anywhere in a program, for example:

const int nrows = 10;

const int ncolumns = 20;

float f[nrows][ncolumns];

However, the array will have to be initialized with executable statements. Also, in contrastto vectors, arrays are not automatically initialized to zero. It must be done either on the dec-laration statement or explicitly initialized to zero within executable statements. Otherwise,the array will contain bit patterns that were left in memory following previous use.

8.1. ARRAYS 203

Here’s a small program that demonstrates some of the declaration, initialization and ad-dressing concepts.

//File: whereItsAt.cpp

#include <iostream>

int main(void)

{ const int NROWS = 3;


int a[NROWS][NCOLUMNS] = {{1}, {3}, {5}};

cout << a[0][0] << ":" << a[0][1] << "\n";

cout << a[1][0] << ":" << a[1][1] << "\n";

cout << a[2][0] << ":" << a[2][1] << "\n";

cout << "&a[2][1]: " << &a[2][1] << "\n";

cout << "&a[2][0]: " << &a[2][0] << "\n";

cout << "&a[1][1]: " << &a[1][1] << "\n";

cout << "&a[1][0]: " << &a[1][0] << "\n";

cout << "&a[0][1]: " << &a[0][1] << "\n";

cout << "&a[0][0]: " << &a[0][0] << "\n";

cout << "a: " << a << "\n";

return 0;

}

which produces the following output:

1:0

3:0

5:0

&a[2][1]: 0xbffff72c

&a[2][0]: 0xbffff728

&a[1][1]: 0xbffff724

&a[1][0]: 0xbffff720

&a[0][1]: 0xbffff71c

&a[0][0]: 0xbffff718

a: 0xbffff718

Note how this multi-dimensional array is stored. Computer memory is organized as if it werea one-dimensional array. Thus, a multi-dimensional array has to be unfolded into a “line”


and stored. Different languages do this in different ways! C++’s convention is to store itfrom the outer index to the inner index, reinforcing the “vectors of vectors” concept. In ourtwo-dimensional example this means that all the columns in the first row are stored, thenall the columns of the second row and so forth.

It is an important performance issue when writing programs to be aware of this storage orderand use the arrays in the most efficient manner. For example, the following program :

//File: fastArray.cpp

int main(void)

{ const int SIZE = 1000;

float a[SIZE][SIZE];

for(int k = 1 ; k <= 100 ; k = k + 1)

for(int i = 0 ; i <= SIZE - 1 ; i = i + 1)

for(int j = 0 ; j <= SIZE - 1 ; j = j + 1)

a[i][j] = 1;

return 0;

}

executes almost twice as fast on my computer than slowArray.cpp which is only differentin the ordering of the two for loops:

//File: slowArray.cpp

int main(void)

{ const int SIZE = 1000;

float a[SIZE][SIZE];

for(int k = 1 ; k <= 100 ; k = k + 1)

for(int i = 0 ; i <= SIZE - 1 ; i = i + 1)

for(int j = 0 ; j <= SIZE - 1 ; j = j + 1)

a[j][i] = 1;

return 0;

}

8.1.3 Passing a two-dimensional array to a function

Here is where multi-dimensional arrays differ significantly from one-dimensional arrays.When a function is passed an array it has to be given enough information to know howto move around in the array. Thus, the dimension of all the outer indices (not the inner-most) must be specified. See how this is done in the following example.

8.1. ARRAYS 205

//File: 2d.cpp

#include <iostream>

#include <iomanip>

#include <cmath>

#include <vector>

const int SIZE = 24;

const int MAX_PRINT = 24;

void printArray(float a[][SIZE])

{ if (SIZE <= MAX_PRINT)

{ for(int i = 0; i < SIZE; i = i + 1)

{ for(int j = 0; j < SIZE; j = j + 1)

cout << setw(2) << static_cast<int>(a[i][j]) << " ";

cout << "\n";

}

}

else cout << "Array is too big to print.\n";

return;

}

void productArray(vector<float>& y, float O[][SIZE], vector<float> x)

{ for(int i = 0; i < SIZE; i = i + 1)

for(int j = 0; j < SIZE; j = j + 1)

y[i] = y[i] + O[i][j] * x[j];

return;

}

int main(void)

{ /* Initialize the D array */

float D[SIZE][SIZE] = {0};

for(int i = 0; i < SIZE - 1; i = i + 1)

{ D[i][i] = -1;

D[i][i + 1] = 1;

}

// Initialize the I array

float I[SIZE][SIZE] = {0};

for(int i = 1; i < SIZE; i = i + 1)

for(int j = 0; j <= i; j = j + 1)


I[i][j] = 1;

// Print out D and I arrays

cout << "The \"D\" array:\n"; printArray(D);

cout << "The \"I\" array:\n"; printArray(I);

// Initialize the f and x vectors

vector<float> f(SIZE), x(SIZE);


{ x[i] = -1 + 2.0*i/(SIZE - 1);

f[i] = 1.0/(1.01 - x[i]);

}

// Operate on f with D and normalize

vector<float> d(SIZE);

productArray(d,D,f);

for(int i = 0; i < SIZE - 1; i = i + 1)

d[i] = d[i]/(x[i+1] - x[i]);

// Operate on f with I and normalize

vector<float> F(SIZE);

productArray(F,I,f);


F[i] = F[i]*(x[i] - x[i-1]);

// Print results

cout

<< " x "

<< " f "

<< " d "

<< " F\n";

for(int i = 0 ; i <= SIZE - 1 ; i = i + 1)

cout << setiosflags(ios::fixed)

<< setw(11) << x[i] << " "

<< setw(11) << f[i] << " "

<< setw(11) << d[i] << " "

<< setw(11) << F[i] << "\n";

return 0;

}

The output of this program is:

8.1. ARRAYS 207

The "D" array:-1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

The "I" array:0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 01 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 01 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0


1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

x f d F-1.000000 0.497512 0.258711 0.000000-0.913043 0.520009 0.283216 0.088480-0.826087 0.544637 0.311375 0.135840-0.739130 0.571713 0.343954 0.185554-0.652174 0.601622 0.381930 0.237869-0.565217 0.634833 0.426560 0.293072-0.478261 0.671925 0.479500 0.351500-0.391304 0.713621 0.542947 0.413554-0.304348 0.760834 0.619879 0.479714-0.217391 0.814736 0.714408 0.550560-0.130435 0.876859 0.832347 0.626809-0.043478 0.949237 0.982116 0.7093510.043478 1.034638 1.176305 0.7993200.130435 1.136925 1.434410 0.8981830.217391 1.261657 1.787930 1.0078920.304348 1.417129 2.290509 1.1311200.391304 1.616304 3.039655 1.2716690.478261 1.880621 4.228182 1.4352010.565217 2.248289 6.283189 1.6307040.652174 2.794654 10.317343 1.8737180.739130 3.691814 20.073689 2.1947440.826087 5.437352 56.080322 2.6675570.913043 10.313904 1031.390381 3.5644211.000000 100.000000 0.000000 12.260068

What does this program do?

Using the following formula for the product of a two-dimensional array with a vector

yi = Ai0x0 +Ai1x1 +Ai2x2 · · ·

we see that the “D” array differentiates and the “I” array integrates the function f(x). Yes,you can do calculus with arrays as long as the spacing of x is such that the function f(x) doesnot change too much from element to element. This is the basis of most calculations thatsolve differential equations. (You won’t be tested on the information in this last paragraph!)

This is what the result looks like:

8.2. CHARACTER ARRAYS 209

−1.0 −0.5 0.0 0.5 1.010−3

10−2

10−1

100

101

102

103

fD*fI*f

8.2 Character arrays

8.2.1 One dimensional character arrays

• We have seen previously that the string class is really a vector in disguise. Onecan also string together characters in a more basic way using one-dimensional arrays.One should only use this form when it is inconvenient to use the string class, asin the example we have just been discussing, because there are some special rulesto remember. Moreover, forgetting these rules can lead to problems related to over-running the boundaries of arrays. Indeed, some of these subtle behaviors are exploitedby destructive computer hackers (the scourge of the computer industry) to break intocomputers and cause damage.

• So, a character string may be represented as a one-dimensional array of single charactersthat terminates with a specific character sentinel, the “null” character, i.e. ’\0’. See!There is a use for something that represents nothing! So, we always have to rememberto include this in the one-character array. If we forget to do this and try to interpret aone-dimensional array of characters as a string, then the code, as it is operating on the


string, will look for the end-of-string sentinel until it finds it, possibly reading (or moredangerously changing) memory locations that were not intended to be accessed. (Thisis a major destructive hacking tool! Don’t do it!) The string class handles all thisautomatically for you, so that is why it is preferable to use the string class wheneverpossible.

• The following declares and initializes a string:

char s[4];

s[0] = ’H’;

s[1] = ’i’;

s[2] = ’\0’; /* null */

Despite the fact that we have allotted 4 bytes for the string, we have only used 3 ofthem because that’s where we put the ’\0’ terminator for that string. After the abovestatements s[3] is undefined.

• Note that the statements

s[2] = ’\0’; /* null */

s[2] = 0; /* the same thing */

are exactly the same thing.

• Strings define their own length with the ’\0’ terminator which acts as a “sentinel” forcompletion of the string.

• A simpler way of initializing a character string at the declaration stage is:

char s[] = {’H’,’i’,’\0’};

which is more convenient than the previous one because there is no need to count thenumber of array elements.

• A simpler way still is:

char s[] = "Hi";

which uses the doubles quotes " to enclose what is called a string literal or a stringconstant. The above array s[] has length 3. The terminating sentinel ’\0’ is put inautomatically by the compiler.

• It is a common mistake to forget about the terminating sentinel ’\0’. For example,say we defined a character string called mistake in the following:

8.3. STRUCTURES 211

char mistake[] = {’b’,’o’,’o’,’b’,’o’,’o’};

As far as C++ is concerned, when the character array mistake[] is accessed as astring, this is not considered to be a string whose value is "booboo"! C++ considersthis to be a string "booboo........\0", that is, it keeps going until it runs into theterminating sentinel ’\0’, or it runs out of memory. This is an error that can easilycrash a program or even worse, cause it to behave erratically.

Can you guess how the following program works?

//File: charArray.cpp

#include<iostream>

#include<string>

int main(void)

{ char correct[12] =

{’I’, ’A’, ’m’, ’C’, ’o’, ’r’, ’r’, ’e’, ’c’, ’t’, ’!’, ’\0’};

char moreData[] = "Will this print?";

char someData[5] = {100, 97, 114, 110, 0};

char mistake[8] = {’A’, ’m’, ’i’, ’s’, ’t’, ’a’, ’k’, ’e’};

cout << mistake << "\n" << correct << "\n";

return 0;

}

8.2.2 Two dimensional character arrays

One can make arrays out of char’s. For example, a checkerboard could be represented by:

char checkerboard[8][8];

for (int i = 0; i < 8; i = i + 1)

for(int j = 0; j < 8; j = j + 1)

checkerboard[i][j] = ’ ’; //Initialize with spaces

checkerboard[0][0] = ’R’; //Place a red checker on the board

checkerboard[7][0] = ’B’; //Place a black checker on the board

8.3 Structures

So far we have been introduced to many forms of variables that contain only one item ofinformation, bool’s, char’s, int’s, float’s and double’s. These are fundamental variable


types in C++, useful interpretations of bits and bytes that are pre-defined in the definitionof the C++ language.

We have also seen more complex objects—classes like vector’s and string’s.

We have also seen arrays of one and two and more dimensions that are made up of only oneof the above fundamental variables types but contain many items referred to by indexingwithin the array.

In many cases, these pre-defined variable types are insufficient to elegantly formulate asolution to an interesting problem. Fortunately, C++ gives the programmer the ability tomake new compound variable types made up of a collection of the existing variable types.This compound variable type is called a structure.

For example, we could define a structure representing a person and the elements of thestructure could be a string for the person’s name, an int for the person’s age, float’s forthe person’s height and weight, more strings for the person’s address and occupation andso on. Structures are the essential composite data type for databases, which represent arelatively large component of the software industry.

Another example relates to the car racing application of Assignment 6. Let’s say that wewanted to introduce new, racing-like features into the game, like oil slicks that reduce tractionon the road surface, or the ability to draft behind the opponent’s car. Every grid pointrepresenting a part of the race track would have to carry more information, the coefficientof friction on the surface, and the direction and number of seconds past the time that a carpassed through it (so that the wind speed could be calculated). While these examples couldbe done by separating the information into different variables, why not put them togetherin a compound variable? After all, they are all related to one other.

8.3.1 The structure definition

An example of a structure definition:

// Define a structure called Student

struct Student

{ string name;

int section;

float grade;

};

The keyword struct begins a structure definition. The identifier Student is called thestructure tag. It is important to note that this is a definition and not a declaration. All weare doing at this point is informing the compiler that we will be declaring variable types ofthe form Student later on in the program.

8.3. STRUCTURES 213

8.3.2 Where can structures be defined?

struct’s can be defined everywhere although it is conventional to define them outside of anyprogramming unit such as main or another function so that all programming units will haveaccess to this variable type.

8.3.3 The members of a structure

The variables within the braces {}’s are the members of the structure called Student. Inthis case the members defined, as suggested by the naming of the member types, are a stringthat will contain a name of a student, an int that will contain the section number of thestudent and a float that will contain the student’s grade. The use of a struct seems quitenatural in the application because the various things associated with any given student canbe variables of various types.

8.3.4 Declaring structure variables

The structure tag is used with the keyword struct to declare (and optionally initialize)variables of the structure type. For example, if we had defined the above structure typeStudent, in the main routine or elsewhere we could declare and assign as follows:

struct Student

Dan = {"Smith, Dan", 210, 79.8},

Jan = {"Brown, Jan", 201, 89.3};

declares two structures of the Student type called Dan and Jan and assigns them names,section numbers and grades.

8.3.5 Assigning values to structure members

Individual structure members can be assigned values by using the “.” notation of the formstructureName.structureMember. For example, if we had not made the assignment in thedeclaration statement, we could have assigned values as follows:

Dan.name = "Smith, Dan"; Dan.section = 210; Dan.grade = 79.8

Jan.name = "Brown, Jan"; Jan.section = 201; Jan.grade = 89.3


8.3.6 Re-assigning values of structure members

Once defined and declared, we can treat these as regular variables, reassigning them, print-ing them, manipulating them in various ways. For example, we could make the followingredefinitions:

Jan.name = "Smith, Jan"; // Jan and Dan get married!

Jan.section = 210; // Jan moves over to Dan’s section (bad idea)

We can also treat a whole structure as a single variable. For example,

Dan = Jan;

reassigns the entire “Dan” structure, giving each member the values of the “Jan” structure.

Consult the complete working code called structure0.cpp that manipulates the “Dan” and“Jan” structures.

8.3.7 Function call-by-value of a structure

Structures can be called by value. This means that the structures are copied locally in thecalled function and the structures employed by the calling routine remain untouched. Hereis an example called structure1.cpp:

//File: structure1.cpp

#include<iostream>

#include<string>

struct Student{string name; int section; float grade;};

void switchStruct(struct Student a, struct Student b)

{ // I/O statements left out

struct Student temp = a; a = b; b = temp; // Switch a and b

// I/O statements left out

return;

}

int main(void)

{ struct Student

Dan = {"Smith, Dan", 210, 79.8},

Jan = {"Brown, Jan", 201, 89.3};

8.3. STRUCTURES 215


switchStruct(Dan,Jan);


return 0;

}

8.3.8 Function call-by-reference to a structure

The more conventional way is to use call by reference for structures, saving memory. Here isthe same example as structure1.cpp but using called by reference. It is called structure2.cpp


#include<iostream>

#include<string>

struct Student{string name; int section; float grade;};

void switchStruct(struct Student& a, struct Student& b)

{ // I/O statements left out

struct Student temp = a; a = b; b = temp; // Switch a and b


return;

}

int main(void)

{ struct Student

Dan = {"Smith, Dan", 210, 79.8},

Jan = {"Brown, Jan", 201, 89.3};


switchStruct(Dan,Jan);


return 0;

}

8.3.9 Structure arrays

One of the most common uses of struct’s is to make arrays of them. Let’s now define anarray of struct’s of the form Student and make each element of the array represent oneof the students of the ENG101 course. Here is how it would be done. The code is calledstructure3.cpp. It creates a class of fictitious students, assigns section numbers and grades.Then it sorts them by name, section or grade, whatever the user wants.



#include<iostream>

#include<iomanip>

#include<string>

#include<vector>

#include<cstdlib>

// Define a structure called Student

const int maxNameLength = 10;

struct Student

{ char name[maxNameLength];

int section;

float grade;

};

bool switchStruct(struct Student& a, struct Student& b)

{ struct Student temp = a; a = b; b = temp; // Switch a and b

return false;

}

void sortStruct(vector<struct Student>& list, bool n, bool s, bool g)

{ bool sorted;

if (n)

{ do

{ sorted = true;

for (int i = 0; i < list.size() - 1; i = i + 1)

{ string string1 = list[i].name;

string string2 = list[i+1].name;

if (string1 > string2)

sorted = switchStruct(list[i],list[i+1]);

}

}while (!sorted);

}

else if (s)

{ do

{ sorted = true;


if (list[i].section > list[i+1].section)


}while (!sorted);

}

else if (g)

8.3. STRUCTURES 217

{ do

{ sorted = true;


if (list[i].grade < list[i+1].grade)


}while (!sorted);

}

return;

}

int main(void)

{ cout << "How many students in the class? ";

int nStuds;

cin >> nStuds;

vector<struct Student> myClass(nStuds);

for (int i = 0; i < nStuds; i = i + 1)

{ myClass[i].section = 200 + rand()%10;

myClass[i].grade = 0.1*(rand()%1001);

int nameLength = 2 + rand()%(maxNameLength - 2);

for (int j = 0; j < nameLength; j = j + 1)

myClass[i].name[j] = 97 + rand()%26;

myClass[i].name[nameLength - 1] = ’\0’;

myClass[i].name[0] = myClass[i].name[0] - 32;

cout << setw(maxNameLength) << myClass[i].name << ", "

<< myClass[i].section << ", "

<< myClass[i].grade << "\n";

}

bool sortByName = false;

bool sortBySection = false;

bool sortByGrade = false;

cout << "Sort by name (0 (no) or 1 (yes))? ";

cin >> sortByName;

if (!sortByName)

{ cout << "Sort by section (0 (no) or 1 (yes))? ";

cin >> sortBySection;

if (!sortBySection)

{ cout << "Sort by grade (0 (no) or 1 (yes))? ";

cin >> sortByGrade;

}


}

sortStruct(myClass, sortByName, sortBySection, sortByGrade);

for (int i = 0; i < nStuds; i = i + 1)

cout << setw(maxNameLength) << myClass[i].name << ", "

<< myClass[i].section << ", "

<< myClass[i].grade << "\n";

return 0;

}

8.3.10 Example using structures and arrays: Ion transport

The following code will be described in class:

//File: survivor.cpp

#include <iostream>

#include <fstream>

#include <iomanip>

#include <cstdlib>

const int NX = 20; // x-grid dimensions

const int NY = 7; // y-grid dimensions

const int STEPS = NY; // # of time steps

const int NCELL = 1000; // # of pairs per cell

const int TOTAL = (NX*NY*NCELL); // total # of pairs

struct N {int nE; int nP;}; // # electrons, # ions

ofstream resultsOnFile; // Open an output file stream

//--------------------------------------------------------------------

//---------------------- Function definitions -----------------------

//--------------------------------------------------------------------

// This function initializes the number of e-’s and ions in each cell

void init(struct N grid[][NY])

{ for (int i = 0; i < NX; i = i + 1)

{ for (int j = 0; j < NY; j = j + 1)

{ grid[i][j].nE = NCELL;

grid[i][j].nP = NCELL;

8.3. STRUCTURES 219

}

}

return;

}

//--------------------------------------------------------------------

// This function counts the number of e-’s

int countE(struct N grid[][NY])

{ int count = 0;

for (int i = 0; i < NX; i = i + 1)

for (int j = 0; j < NY; j = j + 1)

count = count + grid[i][j].nE;

return count;

}

//--------------------------------------------------------------------

// This function counts the number of positive ions

int countP(struct N grid[][NY])

{ int count = 0;

for (int i = 0; i < NX; i = i + 1)

for (int j = 0; j < NY; j = j + 1)

count = count + grid[i][j].nP;

return count;

}

//--------------------------------------------------------------------

// This function recombines the e-’s and ions

void recombine(struct N grid[][NY], int step)

{ for (int i = 0; i < NX; i = i + 1) // Loop over all rows

{ for (int j = step; j < NY; j = j + 1) // Loop over all columns

{ int nE = grid[i][j].nE;

for (int k = 1; k <= nE; k = k + 1) // Loop over all e’s

{ for (int l = 1; l <= grid[i][j].nP; l = l + 1)

{ // Potential annihilation with every ion in the cell

if (0 == rand()%2000) // Small chance of annihilation

{ // If there is an annihilation, reduce numbers by 1

grid[i][j].nE = grid[i][j].nE - 1;

grid[i][j].nP = grid[i][j].nP - 1;

break; // Break out of loop, only one life to give!

}

}

}


}

}

return;

}

//--------------------------------------------------------------------

/* This function shifts the electrons to the right and counts how

many are collected. Note that, depending on the step, we can start

at the y index where the number of e-’s is non-zero.

*/

int tStep(struct N grid[][NY], int step)

{ int escaped = 0;

for (int i = 0; i < NX; i = i + 1)

{ escaped = escaped + grid[i][NY-1].nE;

for (int j = NY - 2; j >= step; j = j - 1)

grid[i][j+1].nE = grid[i][j].nE;

grid[i][step].nE = 0;

}

return escaped;

}

//--------------------------------------------------------------------

// This function prints out the grid

void fGrid(struct N grid[][NY],int step){

resultsOnFile << "Step: " << step << "\n";

for (int i = 0; i <= NX - 1; i = i + 1)

{ for (int j = 0; j <= NY - 1; j = j + 1)

resultsOnFile << "|" << setw(5) << grid[i][j].nE

<< "," << setw(5) << grid[i][j].nP;

resultsOnFile << "|\n";

}

return;

}

//--------------------------------------------------------------------

//--------------------------------------------------------------------

//--------------------------- Main routine ---------------------------

//--------------------------------------------------------------------

int main(void)

{ struct N grid[NX][NY]; // grid is an N struct

8.3. STRUCTURES 221

init(grid); // Initialize the grid

// Output starting conditions

cout << "Step " << setw(2) << 0

<< " : nE = " << setw(6) << countE(grid)

<< " : nP = " << setw(6) << countP(grid)

<< " : nEsc = " << setw(6) << 0 << "\n";

resultsOnFile.open("N.dat");

fGrid(grid, 0); // Write the grid to file

// Loop over time steps, set = to NY grid size

int tEsc = 0; // Total number of escapees

for (int step = 0; step < STEPS; step = step + 1)

{ recombine(grid, step); // Recombine e- & ions

int nEsc = tStep(grid, step); // Take a time step

cout << "Step " << setw(2) << step + 1

<< " : nE = " << setw(6) << countE(grid)

<< " : nP = " << setw(6) << countP(grid)

<< " : nEsc = " << setw(6) << nEsc << "\n";

fGrid(grid, step + 1); // Write grid, next step

tEsc = tEsc + nEsc; // Sum total escapes

}

resultsOnFile.close(); // Close the file

cout << "Chance of survival = " << setprecision(3)

<< static_cast<float>(tEsc)/TOTAL << "\n";

return 0;

}

//--------------------------------------------------------------------


8.4 Problems

1. Multiplying vectors by arrays

Complete the following program. The main routine declares an int array a[ROWS][COLS].Your program should work for any values for ROWS and COLS, not just the ones pro-vided. Read through main() to understand what it is doing. The function print()

prints the array to cout. The function randomFill() fills the array with random int’sbetween 0 and 9. Hint: Use rand()%10. The function mMult() returns a vector whosevalue of its i’th element is:

y[i] = a[i][0]*x[0] + a[i][1]*x[1] + a[i][2]*x[2] + ...

+ a[i][NCOLS - 1]*x[NCOLS - 1];

Hint: Use for loops to do this. Supply the parameter lists for the three functions.Answer the following question:Why is the return vector always full of zeros?

#include <iostream>

#include <vector>

#include <cstdlib>


const int ROWS = 5; // Set to 5 but can be anything

const int COLS = 4; // Set to 4 but can be anything

void print(/* Supply parameter list*/ )

{//Supply code for this function...

}

void randomFill(/* Supply parameter list*/ )


}

vector<int> mMult(/* Supply parameter list*/ )


8.4. PROBLEMS 223

}

int main(void)

{ int a[ROWS][COLS]; // Declare the array

randomFill(a); // Fills array with rand()%10 int’s

print(a); // Prints the array

vector<int> x(COLS); // Declare a vector

// Multiply the vector x by the array a, returning y

vector<int> y = mMult(a,x);

// Print the result

for (int i = 0; i < ROWS; i = i + 1) cout << y[i] << endl;

return 0;

}

2. Making arrays of various types

Consider the following main routine:

int main(void)

{ int I[SIZE][SIZE] = {0}; ident(I); print(I);

int U[SIZE][SIZE] = {0}; upper(U); print(U);

int L[SIZE][SIZE] = {0}; lower(L); print(L);

int H[SIZE][SIZE] = {0}; hatch(H); print(H);

return 0;

}

The 5 functions called by main() produce the output shown following, when SIZE =

10. Write the 5 functions that complete this program. Your program has to work forany SIZE greater than 0.

(Output from the completed program, when SIZE = 10)

1 0 0 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 1 0 0


0 0 0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 0 0 1

1 1 1 1 1 1 1 1 1 1

0 1 1 1 1 1 1 1 1 1

0 0 1 1 1 1 1 1 1 1

0 0 0 1 1 1 1 1 1 1

0 0 0 0 1 1 1 1 1 1

0 0 0 0 0 1 1 1 1 1

0 0 0 0 0 0 1 1 1 1

0 0 0 0 0 0 0 1 1 1

0 0 0 0 0 0 0 0 1 1

0 0 0 0 0 0 0 0 0 1

1 0 0 0 0 0 0 0 0 0

1 1 0 0 0 0 0 0 0 0

1 1 1 0 0 0 0 0 0 0

1 1 1 1 0 0 0 0 0 0

1 1 1 1 1 0 0 0 0 0

1 1 1 1 1 1 0 0 0 0

1 1 1 1 1 1 1 0 0 0

1 1 1 1 1 1 1 1 0 0

1 1 1 1 1 1 1 1 1 0

1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1

1 0 1 0 1 0 1 0 1 0

1 1 1 1 1 1 1 1 1 1

1 0 1 0 1 0 1 0 1 0

1 1 1 1 1 1 1 1 1 1

1 0 1 0 1 0 1 0 1 0

1 1 1 1 1 1 1 1 1 1

1 0 1 0 1 0 1 0 1 0

1 1 1 1 1 1 1 1 1 1

1 0 1 0 1 0 1 0 1 0

3. Vectors and arrays

Complete the program started below.

The main routine declares an int array a[ROWS][COLS]. Your program should work forany values for ROWS and COLS, not just the ones provided. Read through main() tounderstand what it is doing. The function randomFill() fills the array with randomint’s between 0 and 3 inclusive. Hint: Use rand()%4. The function sumRowsAndCols()

8.4. PROBLEMS 225

sums the rows and columns putting the results into two separate vectors, sumRows, avector ROWS in length containing the sum of each row, and sumCols, a vector COLS inlength containing the sum of each column: The function print() prints the array tocout and the vectors as shown in the example below. Note, pretty formatting is notimportant, but the numbers have to be correct. Supply the parameter lists for thethree functions.

Example use:

> a.out

2 2 1 3 3 : 11

3 2 3 0 2 : 10

1 3 0 0 1 : 5

3 1 3 1 2 : 10

- - - - -

9 8 7 4 8

#include <iostream>

#include <vector>

#include <cstdlib>


const int ROWS = 4; // Set to 4 but can be anything

const int COLS = 5; // Set to 5 but can be anything

int main(void)

{ int a[ROWS][COLS];

randomFill(a);

vector<int> sumRows(ROWS);

vector<int> sumCols(COLS);

sumRowsAndCols(a,sumRows,sumCols);

print(a,sumRows,sumCols);

return 0;

}

4. More vectors and arrays

Consider the following main routine:

const int ROWS = 10; //Constant defined as global


const int COLS = 8; //Constant defined as global

int main(void)

{ int a[ROWS][COLS] = {0};

vector<int> r(ROWS);

for(int i = 0; i < r.size(); i = i + 1) r[i] = rand()%4; // 0 <= r[i] <= 3

vector<int> c(COLS);

for(int i = 0; i < c.size(); i = i + 1) c[i] = rand()%4; // 0 <= c[i] <= 3

outerProduct(a,r,c);

print(a,r,c);

return 0;

}

Write the two functions outerProduct() and print(). outerProduct() computesthe outer product of two vectors, as follows:a[i][j] = r[i]*c[j];

where r[i] and c[j] are vectors. print() prints out the vectors and array as shownbelow. You do not have to concern yourself with pretty formatting, but the numbersshould come out in the order shown. Your program has to work for any positive valuesfor ROWS and COLS.

Here is how it works:

red% a.out

v: 1 0 2 2 3 2 1 2

----------------

r[0]: 3| 3 0 6 6 9 6 3 6

r[1]: 1| 1 0 2 2 3 2 1 2

r[2]: 0| 0 0 0 0 0 0 0 0

r[3]: 3| 3 0 6 6 9 6 3 6

r[4]: 1| 1 0 2 2 3 2 1 2

r[5]: 3| 3 0 6 6 9 6 3 6

r[6]: 3| 3 0 6 6 9 6 3 6

r[7]: 1| 1 0 2 2 3 2 1 2

r[8]: 1| 1 0 2 2 3 2 1 2

r[9]: 1| 1 0 2 2 3 2 1 2

5. Complex struct’s

This problem uses a struct defined as follows:

8.4. PROBLEMS 227

struct Complex {double real; double imag;};

We may use struct Complex’s to define “complex numbers”. A complex number isreally a pair of numbers, the “real” part and the “imaginary” part, as suggested bythe above struct definition. If a = (aR, aI) and b = (bR, bI) are two complex numbers,their product is also a complex number ab = (aRbR−aIbI , aRbI+aIbR) Write a functionthat accepts two called-by-reference vector<struct Complex>’s, both of the same sizeand returns a vector<struct Complex> that contains the product according to theabove rule.

6. Using structure lists

Consider the following structure definition:

struct Student {char name[10]; int section; float grade;};

Write a function compatible with the following function call:

adjustGrades(list);

where list is a vector<struct Student> with size > 0. Your function will do thefollowing:

(a) Raise all the grades in the class 5% to a maximum of 100.

(b) Give everyone in the class whose name starts with “Al” a grade of 100.

(c) Sort the list by grades, highest to lowest.


8.5 Projects

1. Battleship! Sink this one!

You will write C++ code that will allow you to play the game “Battleship!” on yourcomputer following the rules given below.

The game is played on a 10 by 10 array of char’s. A battleship is represented by a seriesof sizeShip contiguous (joined) points that lie in a horizontal, vertical or diagonal line.This can be represented by some special character in a 10 by 10 character array, say ’S’.For example, if the array everywhere was initialized to the character ’.’ and you wrote,board[2][3] = ’S’; board[3][4] = ’S’; board[4][5] = ’S’; board[5][6] = ’S’;

it would represent a battleship on a diagonal occupying the above indices.

For each ship you will ask the computer to generate sizeShip such contiguous pointson a line starting from a random position within the entire array. For example, if you:#include<cstdlib>

in the preprocessor area, the code:i = rand()%10; j = rand()%10; board[i][j] = ’S’;

picks a random point in the array called board and assigns it the value S, indicatingthat it is one of the points representing a ship. This is the first point on the ship.The hard part is to pick sizeShip - 1 more points on a line that are contiguous withthe first point and that remain within the array. This you have to figure out on yourown. The orientation of the battleship should be random as well. You have to makethe part of the program that determines the ship’s position and orientation a separateC++ function and the array name must be an argument (in the parameter list) of thisfunction. When you come to the part where you are putting more than 1 ship on theboard, it gets even harder because no 2 ships can occupy the same space.

Now comes the fun part. You (the user of the program) have 50 + shipSize*nShips/2

“bombs” to toss at the battleship, where nShips is the number of enemy ships. Thebattleship position and orientation are random and are kept secret from you until thecode finishes, so you do not know where it is while you are playing. You have to dropthe bomb on the array, which is equivalent to selecting an element of the array, byasking the user to provide two indices that lie within the board. If that array elementis an S, you have hit the ship and the program informs you that you have hit the ship.(See the example below.) You have to hit all locations of a ship to sink it and you haveto sink every ship to win the game.

If you hit a ship once, the game strategy simplifies. You look for adjacent points (8possibilities). If you hit it again, you know the orientation. Now, all you have to do ismake sure you get all the points that lie on the line.

Win or lose, print out a representation of the board that shows clearly: the locationof the ship, the points of the ship that were struck, and the points where bombs weredropped but missed.

8.5. PROJECTS 229

Here is an example of a typical play of the game. Your computer output does not haveto look exactly as follows. However, you have to follow the above rules explicitly.

Number of enemy ships? 1

Size of the enemy ships? (<= 10) 4

Bombs away # 1: i j: 3 4






Bombs away # 7: i j: 2 6 Ouch, that hurts!








Ugh, ya got me! Glub, glub, glub...

. . . . . . . . . .

. . . . . . . . . .

. . . . . o * o . .

. . . . o * o o . .

. o . . * . . . . .

. . . * . . . . . .

o . . . . . . . . .

. . o . . . . . . .

. . . . . . . . . .

. . o . . . . . . .

A "." is "open sea".

An "*" is a "hit on an enemy ship".

An "S" is a "surviving enemy ship".

An "o" is a "missed bomb".


2. Ladies and Gentlemen: Start your engines!

Write a computer program that simulates a race car negotiating an S-curve track andeither crashing or reaching the finish line. The game is played by either one playerracing against the clock, or two players competing against each other.

The game starts with a printout of a 2-D grid representing the track made up of a two-dimensional array of characters (char’s) with 52 rows and 72 columns. The picture onthe following page represents the starting condition for a one-person game. The grid canbe printed to the terminal screen by writing the entire char grid to cout. If you havetrouble seeing the whole grid in a single window, try using <Control>-<right mouse button>

in an XTerm window and resizing the window to the whole screen.

The outer boundary of the track and the two impenetrable barriers are represented by’X’’s, the open track by spaces ’ ’, the finish line by ’F’’s (lower right hand corner)and the car, pictured at its starting location, grid[1][1] (upper left hand corner), byan ’O’. In the following picture, the rows go down the page top to bottom and thecolumns go left to right.

Here’s how the game is played. The user is continually prompted (until the gameends) for a component of horizontal (positive is left to right) and vertical acceleration(positive is top to bottom). (You imagine that this prompt is given once per second,that is, the user has a chance to change the acceleration once/second.) The userresponds with either -1, 0, or 1 for each of these, effectively steering, accelerating andbraking the car. If the user responds with a number outside of this range, the carcrashes and the game is over, because either the engine explodes or the brakes andtires burn up—real concerns for real race cars.

The velocity in both the vertical and horizontal directions starts at 0 and increases ordecreases after each prompt, for example, xVelocity = xVelocity + xAcceleration,and the same for the y direction. Then, both components of the position are changed,for example, xPosition = xPosition + xVelocity, and the same for the y direc-tion. A car crashes when its position takes it outside the racetrack (except when itcrosses the finish line) or into a wall or barrier, or an improper acceleration was input.Crossing the finish line “wins” the race.

8.5. PROJECTS 231

The starting grid in a one-person game (note prompt for acceleration at bottom):

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

XO XXXXXXXXXXXXXXXXXXXX X

X XXXXXXXXXXXXXXXXXXXX X














X XXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXX X




















X XXXXXXXXXXXXXXXXXXXXXXXXX X















XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXFFFFFFX

Horizontal and vertical acceleration (-1,0,1):

To start the programming, write a working main routine and a function called initRacethat initializes the grid to the given starting configuration, and a function calledprintRace that prints the grid. The walls and barriers have to be exactly as shownin the figure on the last page. The barrier on the left is 20 columns wide and 35 rowshigh. The barrier on the right is 25 columns wide and 35 rows high.


Then, program a one-person and a two-person game. Keep track of the total time foreach player. If a player crashes he is out of the race. The first across the finish linewins! Hint: Program a one-person game first, then adapt your program for two people.

Warning: This game is addictive!

Consider modifying the program to do one of the following. These are quite challengingbut well within your capabilities.

• In a one-person game, keep track of the real time it takes to respond and adjustthe algorithm to work with real time. (Use the standard time library ctime. Youwill have to look up the standard time functions in a different book.)

• Program a one-person-against-the-computer game with multiple plays. Initially,the computer makes random moves. Once the player has crossed the finish line,the computer “learns” the players solution and tries to improve it.

• Program some nice graphics for this game. (Good opportunity to read ahead inMatlab.)

• Program a two-person game but one that generates many barriers randomly in agame with very many columns that scrolls down the screen.

• Dream up some other nice variant of the game.

Sample plays of the race on the following pages

8.5. PROJECTS 233

User input: (0,1),(0,1)(0,1),(0,1),(0,1),(1,1),(1,1),(1,1),(1,1),(1,1)

















XO XXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXX X






X O XXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXX X







X O XXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXX X








X O XXXXXXXXXXXXXXXXXXXXXXXXX X









X O XXXXXXXXXXXXXXXXXXXXXXXXX X





XXXXXXXXXXXXXXXXOXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXFFFFFFX

Crashed after 10 seconds Note the car buried in the wall! (Ouch!)


Almost. Lost control at the very end.


XO XXXXXXXXXXXXXXXXXXXX O O X










XO XXXXXXXXXXXXXXXXXXXX O OX





XO XXXXXXXXXXXXXXXXXXXX O XXXXXXXXXXXXXXXXXXXXXXXXX O






X O XXXXXXXXXXXXXXXXXXXX O XXXXXXXXXXXXXXXXXXXXXXXXX X


















X O O XXXXXXXXXXXXXXXXXXXXXXXXX X











XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXFFFFFFX

Crashed after 33 seconds

8.5. PROJECTS 235

Richard Petty, eat your heart out!(Solution by Ali Mehmet Kutman, Fall 2000 ENG101/CoE student at UoM)













X XXXXXXXXXXXXXXXXXXXX O O X




XO XXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXX X


X XXXXXXXXXXXXXXXXXXXX OXXXXXXXXXXXXXXXXXXXXXXXXXO X




XO XXXXXXXXXXXXXXXXXXXX O XXXXXXXXXXXXXXXXXXXXXXXXX O X





X O XXXXXXXXXXXXXXXXXXXX O XXXXXXXXXXXXXXXXXXXXXXXXX O X






X XXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXX O X







X XXXXXXXXXXXXXXXXXXXXXXXXX O X








X XXXXXXXXXXXXXXXXXXXXXXXXX O X



Finished after 31 seconds (Can anyone beat Ali’s 31 second solution?)

3. Mastermind: human vs. machine!

For this project you will write a variant of the game called “Mastermind”.

Here is a description of the game:

• The computer picks 4 random numbers between 1 and 9 inclusiveHint: hiddenNumber = rand()%9 + 1


for each of two players, those players being the human (person using the program)and the machine (the computer). Some of these numbers can be the same, or theycan all be different, depending on what rand() returns.

• On each play, the human picks 4 numbers and the machine picks 4 numbers. Ifone of the numbers is correct and in the right position, the hidden number isrevealed.

• The first to reveal all 4 numbers wins the game. There are three possible outcomes1) human wins, 2) computer wins, 3) it’s a tie.

• The playing boards must be represented by integer arrays.

• The algorithm that the machine employs to play the game must be designed so thatthe machine is guaranteed to win in 9 moves, unless the human has already won.Otherwise, you are free to choose whatever algorithm you like. Look at the ex-ample below. It gives a big hint as to how to design a simple algorithm that isguaranteed to work well.

HUMAN MACHINE

===== =======

? ? ? ? ? ? ? ?

------- -------

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

Enter your guess: 1 2 3 4

HUMAN MACHINE

===== =======

? ? ? ? ? ? ? ?

------- -------

1 2 3 4 1 1 1 1

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0


HUMAN MACHINE

===== =======

? ? ? 5 ? 2 ? ?

------- -------

1 2 3 4 1 1 1 1

2 3 4 5 2 2 2 2

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

8.5. PROJECTS 237

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0


HUMAN MACHINE

===== =======

? ? ? 5 ? 2 ? ?

------- -------

1 2 3 4 1 1 1 1

2 3 4 5 2 2 2 2

3 4 5 5 3 2 3 3

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0


HUMAN MACHINE

===== =======

? 5 ? 5 4 2 ? ?

------- -------

1 2 3 4 1 1 1 1

2 3 4 5 2 2 2 2

3 4 5 5 3 2 3 3

4 5 6 5 4 2 4 4

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0


HUMAN MACHINE

===== =======

5 5 7 5 4 2 5 ?

------- -------

1 2 3 4 1 1 1 1

2 3 4 5 2 2 2 2

3 4 5 5 3 2 3 3

4 5 6 5 4 2 4 4

5 5 7 5 4 2 5 5

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

Curses, human, you win!


Chapter 9

Miscellaneous Topics

9.1 Generating random numbers

The C++ standard library, <cstdlib>, provides a way of producing a random number, ac-tually a “pseudo” random number using the rand() function. The rand() function producesa pseudo random integer between 0 and RAND MAX, inclusive. The constant RAND MAX is alsodefined in <cstdlib>.

Here is an example program, called rand0.cpp that calls rand(). Note the inclusion of<cstdlib>.

//File: rand0.cpp

#include <iostream>

#include <iomanip>

#include <cstdlib> // Need this for the rand() function

int main(void)

{ cout << "RAND_MAX = " << RAND_MAX << "\n\n";

cout << "The first 10 random numbers are...\n";

for(int i = 1; i <= 10; i = i + 1)

cout << "Random number "

<< setw(2)

<< i << " = "

<< rand() // Random int between 0 and RAND_MAX inclusive

<< "\n";

return 0;

}

239

240 CHAPTER 9. MISCELLANEOUS TOPICS

Compiling and executing this results in:

RAND_MAX = 2147483647

The first 10 random numbers are...

Random number 1 = 1804289383










Note that RAND MAX and the implementation of rand() is system dependent. You should notexpect the same values on a different computer or even a different version of the operatingsystem on the same computer.

If you run this program again, you will note that the same random numbers are produced!This sort of goes against the idea that rand() produces random numbers! Actually, this isa benefit. Since the behavior is predictable, it makes it easier to debug. Imagine if you weretrying to debug a program that always gave different results! You would go bonkers tryingto find the cause of the error. So, predictable random numbers are good. But, this is whythey are called “pseudo” random numbers. They are made from a mathematical prescriptionthat is remarkably simple. Yet, they pass many different tests of “randomness”. They alsofail some tests as well but not ones that we are likely to try.

If you are not happy with the same random sequence every time, you can “seed” the randomnumber to produce another sequence. The <cstdlib> function srand() does this. All youhave to do is provide a positive integer to the argument list of srand(). Here’s an examplecalled rand1.cpp:

//File: rand1.cpp

#include <iostream>

#include <iomanip>

#include <cstdlib> // Need this for the rand(), srand() functions

int main(void)


cout << "Input any int to seed the random number generator: ";

int seed;

9.1. GENERATING RANDOM NUMBERS 241

cin >> seed;

srand(seed);


for(int i = 1; i <= 10; i = i + 1)


<< setw(2)

<< i << " = "


<< "\n";

return 0;

}

Compiling and executing the above code results in:

RAND_MAX = 2147483647

Input any int to seed the random number generator: 123456789












Note the different sequence because the user provided the seed 123456789 to srand().

One way to make the start-up of a program that uses rand() appear to be different everytime, is to seed the random number generator with the int that is provided by the time()function. The time() function is part of the standard library <ctime>. Here’s an examplecode called rand2.cpp that uses it:

//File: rand2.cpp

#include <iostream>

#include <iomanip>

#include <cstdlib> // Need this for the rand() and srand() functions

#include <ctime> // Need this for the time() function


int main(void)


srand(time(NULL)); // Seed the random number generator


for(int i = 1; i <= 10; i = i + 1)


<< setw(2)

<< i << " = "


<< "\n";

return 0;

}

Try running rand2.cpp and verify that you get different numbers every time. This uniquestart-up feature is very useful for programming games, in order to produce a game that startsdifferently every time.

The interesting uses of rand() involve making decisions based on the output of rand() andthis simulates random behavior. In many applications you can use the modulus function todo this. For example,

card = rand() % 13 + 1

has a one in 13 chance of being a one, a one in 13 chance of being a two, etc. Thus, youcould use the above statement to simulate the random dealing of cards and you could assigna one to an “ace”, a 2 to a “deuce”, 11 to a “jack”, etc. If you were dealing with a finiteseries of cards, however, you would have to take some care that you do not deal, say, 5 acesand that would require some extra care with the coding. We won’t go into further detailhere on this topic except to say that dealing 5 aces can get you into BIG trouble in certaincircumstances.

Another way of using random numbers, more common to scientific disciplines, is to convertthe random integer to a random double or float, usually converted so that the number liesbetween 0 and 1 (may be inclusive or exclusive of one or both of the endpoints).

Here is an example code, called rand3.cpp, that converts the random integer to a doublebetween 0 and 1 inclusive:

//File: rand3.cpp

#include <iostream>

#include <iomanip>

#include <cstdlib> // Need this for the rand() and srand() functions

#include <ctime> // Need this for the time() function


int main(void)



int nRandoms = 10;

cout << "The first " << nRandoms << " random numbers are...\n";

for(int i = 1; i <= nRandoms; i = i + 1)

{ double randomDouble = rand()/static_cast<double>(RAND_MAX);


<< setw(2)

<< i << " = "

<< randomDouble // Random double between 0 and 1 inclusive

<< "\n";

}

return 0;

}

Here’s an example of its use:

RAND_MAX = 2147483647


Random number 1 = 0.174225

Random number 2 = 0.130327

.

.

.

Application: Calculating π by throwing darts

If you are as bad a dart player as I am, throwing a dart at a circular object is truly a randomprocess. Imagine that you have a circle inscribed in a square. It is easy to show that theratio of the area of the circle to the area of the square is π/4. This suggests a way to estimatewhat π is by throwing darts! However, it only works well if you are a bad dart player, likeme.

Set up a circular dartboard on a wall. Draw a square on the wall so that the circle definingthe dartboard touches each side of the square only once. Now start throwing the darts(badly). Ignore all the darts that land outside of the square. Throw a lot of darts! Countthe ones that land on the dartboard and all the ones that land within the square. The ratioshould be close to π/4. And, your estimate gets better the more darts you throw. Warning:


Don’t do this in your dorms. Don’t do this at home. Try it outside against a fenceor something and make sure that other people, pets, your mother’s favorite vase, etc. do notget damaged.

A much safer way is to do it with a computer code. The following code, called randomPi.cpp

calculates π by random sampling:

//File: randomPi.cpp

#include <iostream>

#include <cstdlib> // Need for rand() and srand()

#include <cmath> // Need for pow() and atan()

#include <ctime> // Need for time()

/* Define a conversion function that converts the int returned from

rand() to a double between -1.0 and 1.0 inclusive

*/

#define MYRAND (2.*rand()/RAND_MAX - 1.)

int main(void)

{ int nDarts = 10000; // Number of darts to throw

int nInside = 0; // Counter for those that land inside


for(int i = 1; i <= nDarts; i = i + 1)

{ double xDart = MYRAND;

double yDart = MYRAND;

if (pow(xDart,2) + pow(yDart,2) <= 1.0) nInside = nInside + 1;

}

double pi = 4.0*nInside/nDarts; // Estimate for Pi

double Pi = 4.0*atan(1.0); // This is how you can get a very

// accurate estimate of the real Pi!

cout << "The estimated value of Pi is " << pi << "\n";;

cout << "The true value of Pi is " << Pi << "\n";;

return 0;

}

There are a few new features in this program that you have not seen before.

• The statement#define MYRAND (2.*rand()/RAND_MAX - 1.)


in the pre-processor area replaces every occurrence of MYRAND throughout the rest ofthe file with (2.*rand()/RAND_MAX - 1.). This happens before compilation. It isexactly as if you took an editor, looked for every occurrence of MYRAND and pasted inits replacement (2.*rand()/RAND_MAX - 1.). You can be as creative as you wish with#define ’s, but make sure that you understand this as a cut and pasting procedure.Generally, you should avoid #define ’s because the C++ parser does not help youmuch with syntax errors. But, #define ’s are a part of the language and using themcan make life a lot easier sometimes. Just be careful with them.

• The expression (2.*rand()/RAND_MAX - 1.) produces a random double between -1and 1 inclusive. Why? rand() produces an int between 0 and RAND_MAX, inclusive.Following the rules of precedence and the rules of variable type promotion multiplyingit by “2.” produces a random double between “0.” and “2.*RAND_MAX”. Dividingby RAND_MAX produces a random double between “0.” and “2.”. Finally, subtracting“1.” from it produces a random double between “-1.” and “1.”. There is a lot goingon in this simple-looking expression.

• When you throw a dart on a 2-dimensional board, you are really choosing randomlyan x-coordinate and a y-coordinate. So the statements:double xDart = MYRAND; double yDart = MYRAND;

pick a random point in a square defined by the corners:(x, y) = (−1,−1), (−1, 1), (1, 1), (1,−1).

• The pow() function, part of the standard library <cmath>, takes the power of the firstargument to degree of the second argument. For example, pow(x,3) computes x3.

• So, the statement:if (pow(xDart,2) + pow(yDart,2) <= 1.0) nInside = nInside + 1;

computes x2+y2 and compares it to 1. If x2+y2 ≤ 1, the dart has landed in the circleand it is counted.

• Finally, the standard library math function atan(), also part of <cmath> computesthe arc tangent of its argument, assumed to be in radians. Since tan−1(1) = π/4,4.0*atan(1.0) is a very accurate determination of π, as good as the math library canprovide, at least for a double.

Here’s what 20,000 points tossed at a circle look like, generated by the program above.

246 CHAPTER 9. MISCELLANEOUS TOPICS20000 random points within a squaren(inside circle)/n(total) = π/4

9.1.1 Example: Integrating functions by random sampling

The same idea may be used to integrate functions by random sampling. The following code,called randomIntegrate.cpp integrates exp(−x) for 0 ≤ x ≤ 1.

//File: randomIntegrate.cpp

#include <iostream>

#include <iomanip>

#include <cstdlib> // Need for rand() and srand()

#include <cmath> // Need for exp(), the exponential function

// and abs(), the absolute value function

/* Define a conversion function that converts the int returned from

rand() to a double between 0.0 and 1.0 inclusive

*/

#define MYRAND (static_cast<double>(rand())/RAND_MAX)


int main(void)

{ int nDarts = 1000000000, under = 0;

double Ans = 1 - exp(-1);

for(int i = 1; i <= nDarts; i = i + 1)

{ double xDart = MYRAND;

double yDart = MYRAND;

if (yDart <= exp(-xDart)) under = under + 1;

if (0 == i%1000000) // Every million times

{ double ans = static_cast<double>(under)/i;

cout << setiosflags(ios::fixed) << setprecision(8)

<< "#" << setw(10) << i

<< ": I = " << setw(10) << ans

<< ", Delta = " << abs(1 - exp(-1) - ans) << "\n";

}

}

return 0;

}

There are several new features in this code:

• The standard math library functions defined in <cmath>, exp(), for the exponentialex and abs() for the absolute value, are used.

• setiosflags(ios::fixed) allows floating point numbers to print with trailing zeros,nice for formatting. This only has to be done once.

• setw() sets the width of the output of the following item. It has to be set every time!setw() is defined in <iomanip>.

• setprecision() set the precision of the following number. setprecision() is alsodefined in <iomanip>.

The output of the above program looks like:

# 1000000: I = 0.63213700, Delta = 0.00001644

# 2000000: I = 0.63193800, Delta = 0.00018256

# 3000000: I = 0.63197933, Delta = 0.00014123

# 4000000: I = 0.63180525, Delta = 0.00031531

# 5000000: I = 0.63187300, Delta = 0.00024756

.

.

.


The program operates very much like the program that found π by random sampling. Wefind an x and a y randomly. However, the only difference is that we count only those pointsthat fall under the function as contributing to the area under the curve. The area under thecurve is the fraction of points that fall under the curve, times the area of the rectangle. Inthe above example, the area of the rectangle is 1. If the rectangle did not have unit area,how would you know what to multiply by? That’s what you have to figure out in assignment4!

9.2 Simple Sorting: The bubble or sinking sort

This is called the bubble or sinking sort technique because small items “bubble” to the topand large items “sink” to the bottom. The technique involves looping through an arraylooking at pairs of values in order. If the “upper” item is bigger than the “lower” item, theyare switched. Then the next pair is considered. The “lower” item of the previous pair isthe “upper” item of the current pair. The iteration continues until you can go through theentire array and no pair is switched.

For example, imagine that you have the following array {3, 2, 1} that you want sorted inascending order. We will use the convention that parentheses surround the pair underconsideration.

Here is how it would work:

Loop 1: {3, 2, 1} : {(3, 2), 1} ⇒ {2, (3, 1)} : {2, 3, 1} Change, continueLoop 2: {2, 3, 1} : {(2, 3), 1} ⇒ {2, (3, 1)} : {2, 1, 3} Change, continueLoop 3: {2, 1, 3} : {(2, 1), 3} ⇒ {1, (2, 3)} : {1, 2, 3} Change, continueLoop 4: {1, 2, 3} : {(1, 2), 3} ⇒ {1, (2, 3)} : {1, 2, 3} No change, stop

Here is some “boiler plate” pseudocode that gets the job done:

bool whileThingsAreChanging = true;

while(whileThingsAreChanging)

{ whileThingsAreChanging = false; //Assume that the list is sorted

for(/* appropriate for-loop parameters */) //Careful!

{ //Loop through the list in an order fashion in one direction

//Compare adjacent pairs of things

//If the pair is out of order, switch them and set

// whileThingsAreChanging to true

}

//When the for loop completes, either nothing has changed and

//whileThingsAreChanging has remained false, or something changed and

//whileThingsAreChanging was changed to true

9.3. RECURSION—A FUNCTION CALLING ITSELF 249

}

//When the execution arrives here, the list is sorted.

9.3 Recursion—A function calling itself

Sometimes in mathematics, there are very compact ways of writing rules for generatingcomplicated things. For example, the factorial can be written out explicitly as:

N ! = N × (N − 1)× (N − 2) · · ·3× 2× 1

or, it can be written more compactly as

N ! = N × (N − 1)! ∀N >= 0 where 0! ≡ 1

where N is an integer.

Mathematicians prefer the second expression because it is better defined (no assumptionsabout what · · · means) and it also suggests a mechanism or process for calculating N ! C++mimics this by providing the programmer a mechanism called recursion—the ability for afunction to call itself. Here is how the factorial can be calculated using this technique.

//File: recursiveFactorial.cpp

#include <iostream>

int factorial(int n)

{ if (0 == n) return 1;

else return (n * factorial(n - 1));

}

int main(void)

{ cout << "N? ";

int N;

cin >> N;

cout << N << "! = " << factorial(N) << endl;

return 0;

}

Note how compact the coding is, and how similar it is to the second mathematical definition.Note also how the 0! is coded in the if statement. This is critical to stop the recursion process.Try writing out the stack frames for a simple case, say 5!, to see how this works.

There are a few simple, general guidelines for using recursion:


• Successive calls to functions should represent progressively simpler cases. (For example,(N − 1)! is “simpler” than N !)

• There is an end to the recursion, the “base case”, that does not involve a recursivecall. (For example 0! = 1)

• If a function calls itself more than once, and the “nesting level” gets deep, the calcu-lation can be very inefficient. (See the following example.)

• Recursion is a great way to crash a computer or a program! If you do not program thestopping criterion correctly, you will cause your stack to collide with your heap, andyour program will be in a heap of trouble.

To illustrate the third point above, consider the calculation of the Fibonacci numbers definedas follows.

f(N) = f(N − 1) + f(N − 2) ∀N > 0 where f(1) ≡ 1 f(2) ≡ 1

where N is an integer. The code can be written using recursive calls as follows:

//File: recursiveFibonacci.cpp

#include <iostream>

int fibonacci(int n)

{ if (1 == n || 2 == n) return 1;

else return (fibonacci(n - 1) + fibonacci(n - 2));

}

int main(void)

{ cout << "N? ";

int N;

cin >> N;

cout << "Fibonacci(" << N << ") = " << fibonacci(N) << endl;

return 0;

}

The fibonacci function calls itself twice. This is a NO, NO! unless N is quite small, likeless than 30. Don’t believe it? Try calculating f(1000)!

Although the above program is elegant and compact, we sometimes have to give up clarity forefficiency, yet another compromise we must sometimes make. Here is how one can programthe Fibonacci numbers using loops.

9.3. RECURSION—A FUNCTION CALLING ITSELF 251

//File: loopFibonacci.cpp

#include <iostream>

int fibonacci(int n)

{ if (1 == n || 2 == n) return 1;

else

{ int fMinus2 = 1;

int fMinus1 = 1;

int f;

for (int i = 3; i <= n; i = i + 1)

{ f = fMinus2 + fMinus1;

fMinus2 = fMinus1;

fMinus1 = f;

}

return f;

}

}

int main(void)

{ cout << "N? ";

int N;

cin >> N;

cout << "Fibonacci(" << N << ") = " << fibonacci(N) << endl;

return 0;

}

It’s ugly but it’s fast. Try comparing the speed of execution of the above two methods. Youwill be surprised at the difference.

Recursion relations play a big role in mathematics and consequently there are many examplesthat could be coded in C++ and used as examples. However, another important use ofrecursion relations relates to graphics. The following example recursively splits a line intotwo smaller segments chosen randomly but with the same total length. This is a very simplemodel of how a tree grows. The output of this program is the total number of branchesaccording to the user’s input of the largest branch length and the smallest branch length.

//File: tree.cpp

#include <iostream>

#include <cstdlib>

int tree(float L, float LMin)

{ if (L <= LMin) return 0;


else

{ float split = static_cast<float>(rand())/RAND_MAX;

return 2 + (tree(split * L, LMin) + tree((1 - split) * L, LMin));

}

}

int main(void)

{ cout << "LTotal, LMin? ";

float LTotal, LMin;

cin >> LTotal >> LMin;

cout << "Number of branches = " << tree(LTotal, LMin) << endl;

return 0;

}

There are two companion programs that will be demonstrated in class that provide graphicaloutput from this program. These are called treeGraphics.cpp and anotherTree.cpp,both posted on the web. In class, the close connection to fractals will be evident from thegraphical output. Fractals are interesting not only mathematically, but have great uses incommunications, encryption and virtual cinematography!

Before closing the section on recursion, it is worthwhile to note that this technique is alsoused in numerical simulation, something the author of this lecture is passionate about. Infact, the coding in the tree programs shares a lot in common with, to cite a few examples,scattering of light in the atmosphere, the movement of neutrons in a nuclear reactor, thescattering of electrons and photons in the human body during diagnostic X-ray proceduresor treatment of cancer with radiotherapy beams. Just a few of the details are different. Inorder to find out what these details are, you will have to take a later course in numericalsimulation.

9.4 Input and Output using files

9.4.1 File and stream handling in C++

Why files?

• Except for a few cases where we generated random data using quasi-random numbers,so far we have been getting data into our programs by typing input at the keyboardand any output that our programs have generated has gone to the computer screen.

• This is really only useful for limited amounts of data, either on the input side or theoutput side.

9.4. INPUT AND OUTPUT USING FILES 253

• A large quantity of data is handled more easily and more reliably with files.

• Moreover, the same data files can be used by more than one program.

How is data organized?

• Computers only know about bits - binary digits, 0’s and 1’s. These are the onlyfundamental data types understood by computer hardware.

• Humans, on the other hand, are more comfortable with characters - and these, as wehave come to know are usually represented by 8 bits, a byte, by computers.

• Characters can be combined into fields—combinations of characters that have someunique meaning, like a floating-point number, an integer, a string of characters repre-senting a person’s name, etc

• Character fields can be gathered into records—of related fields, for example, a person’sname, social security number, age, salary, etc as many fields as are needed for a givenapplication.

• Records can be combined into files—collections of records all somehow related, thatcan be accessed by a computer program for some application.

• Files are collected in directories. A single directory usually contains files with re-lated data. Directories can also contain sub-directories with their own files and sub-directories. It is useful to think of a “tree” structure of directories. Each sub-directoryis another branch of the tree. The root directory is the “trunk” of the tree.

• The disk on a computer can contain several such “trees” or “partitions. The “diskcontroller”, a hardware component in a computer, works with the operating system toorganizing or manage the files on the disk.

• A single computer can contain several disks. Some computers, file servers, can housemany.

• These disks (or portions of them) may be shared across networks with other comput-ers either nearby (Local Area Networks [LAN’s]) or over long distances (Wide AreaNetworks [WAN’s]).

• The ability to connect computers and their data together makes possible the establish-ment of the World Wide Web [WWW], which permits access to some of the data thatresides on a computer connected to the network. This data can be made available toanyone who knows its Uniform Resource Locator [URL]. The URL not only gives theinternet address of the data, but also conveys some information about what form thedata is in.


What is C++’s view of files?

• C++ views a file as a sequential stream of bytes. The end of a file is determinedanywhere by an end-of-file (EOF) marker, or at a specific byte count that the operatingsystem has been told about in a program.

END OF FILE (EOF)

• Files are stored permanently by the operating system, usually on a hard-disk or someother medium (floppies, tape, CD’s, DVD’s). The operating system is responsible forthe maintenance of files.

• When a file is opened by C++ what it really means is that a request for data is sent tothe operating system and the data is put into a data stream—and C++ only concernsitself with handling the stream, not the hardware storage device. Once data entersthe stream its is the responsibility of C++ to interpret the data using formattinginstructions.

• This allows most of C++ to be operating-system independent.

• The standard header files iostream (keyboard and screen-related Input/Output [I/O])and fstream (file-related I/O) contains all the stream-file operating system dependen-cies and has to be rewritten for each new operating system.

• When a C++ program starts with #include <iostream> it knows about three streams:

– stdin (standard input) is (usually) the keyboard which can be read using the“cin >>” function.

– stdout (standard output) is (usually) the screen which can be written to usingthe “cout <<” function.

– stderr (standard error) is (usually) the screen which is written to using the“cerr <<” function. The stderr stream is used by C++ to inform the user, viawarning and error messages, that data may contain ambiguities or errors. It isoften useful to make a distinction between stdout and stderr even though theyoutput, by default, to the same hardware device.


• It is possible in C++ to redirect any or all of the stdin, stdout or stderr files.We will demonstrate how to do this only as required for some specific purpose. It isoperating-system dependent.

• It is possible in C++ to define, create and read other files. Just keep in mind thatC++ creates these as streams and lets the operating system handle all the other detailsof file management.

9.4.2 Creating sequential access files and writing to them

What is a sequential access file?

• A sequential access file is one in which the records do not have a fixed, specified length.

• One usually uses such files for writing and reading data that is not changed later andthat is read all at once.

• It is the most efficient way of storing files which have variable length records.

• Changing the contents of records or the numbers of records can be done, but not veryefficiently.

• Accessing the contents of some specific records can be done, but not very efficiently.

• Nonetheless, it is a useful form for storage of data with variable record length that isall read into memory, manipulated and then all written out.

• Sequential files contain character data that is human readable and transportable (usu-ally) from machine to machine and (usually) from architecture to architecture as longas all the machines understand ASCII character representation. The only essentialdifference is the interpretation of the end of record. UNIX uses “line feed”, MacOSuses “carriage return” and DOS/Win uses “carriage return” followed by “line feed”.

There are 6 things to remember when writing to files

1. #include <fstream> must go in the pre-processor area.

2. Declare an output file object and name it using an identifier in C++. This is doneusing the ofstream keyword. For example,

ofstream myOutputFile;


associates the identifier myOutputFile with a user-declared output file stream. Youshould think of this as a declaration, much as a usual variable declaration, or stringor vector object declaration. It can go anywhere in a program. It is good practice tolocate it close to where you use it for the first time.

3. Before writing to the user-define output stream, it must be “opened” using the .open()output file object member function. For example,

myOutputFile.open("someData.dat");

opens the output file stream object called myOutputFile. It also connects the out-put file identifier myOutputFile with a file called someData.dat. Data going tomyOutputFile will be written physically in the file someData.dat located on the com-puter’s hard disk (usually). In this example, the file someData.dat would reside in thesame subdirectory as the load module that was used to start the program assumingthat the user started the program with a statement like:

> a.out

You can also use relative or absolute path names for the parameter of the .open()

member function. For example, a typical unix system usage would be:

myOutputFile.open("~/Private/someData.dat");

and a Dos/Win usage:

myOutputFile.open("C:\\eng101\\hw\\hw6\\hw6.dat");

Can you guess why there are double backslashes above?

Warning! If the parameter of the .open() function refers to an existing file,that file will be destroyed unless 1) it is write-protected, or 2), you takespecial care to see if it exists before opening it. (More on this later.)

4. Always check to see that the file was opened successfully using the .fail() memberfunction. The .fail() member function returns a true when the file open failed. Forexample,

if (myOutputFile.fail())

{ //File open fail

return 1; // Return to calling function with abnormal termination code.

}


5. Write to the output stream. Remember it is just a stream, just like cout, but withyour own identifier. For example,

int i = 1, j = 2;

myOutputFile << i << ", " << j << endl;

6. When you are finished writing to the file, close the file using .close member function.For example,

myOutputFile.close();

The .close() member function disconnects the output stream object, in this case,myOutputFile, with the physical file, in this case, someData.dat. Although the ter-mination of a program will properly close all files that it opened, it is a good idea toclose files as soon as you are finished with them. This frees up some system resourcesfor other programs and in the event of a system crash, closed files are left intact ormay be recovered more easily. Files left unclosed during a system crash may be leftincomplete. (This is somewhat system dependent and dependent upon how suddenthe crash is. Power failures are particularly nasty in this regard.)

Here is an example called fileOutput.cpp that illustrates all these concepts, plus a fewother concepts.

//File: fileOutput.cpp

#include <iostream>

#include <fstream>


int main(void)

{ ofstream myOutputFile; // Create an output file object

cout << "What filename do you want to write to: ";

char fileName[20]; // 1D array of chars

cin >> fileName;

myOutputFile.open(fileName); //Connect the stream to the file


{ // File could not be opened

cerr << "File called " << fileName << " could not be opened.\n";


return 1; // Return to O/S with abnormal return code

}

else

cout << "File called " << fileName << " was successfully opened.\n";

bool keepReading = true; // Keep reading if true

do{

cout << "Two floats? ";

float x, y; // Two input floats

cin >> x >> y;;

if (cin.eof()) keepReading = false; // End of file <CNTL>-D detected

else myOutputFile << x << " " << y << endl;

}while(keepReading);

cout << "End of input detected\n";

myOutputFile.close(); //Disconnect the stream from the file

return 0;

}

This code makes use of the .eof() member function which tests to see if and end-of-file(EOF) marker has been received in the stream that it is referring to, in this case, cin..eof() returns true if the EOF has been received and false otherwise. Note how this isexploited in the above example. This is how you can control file input with an EOF sentinel.

On UNIX systems, EOF is signaled with a <RETURN><CNTL>-d, on MS systems by <CNTL>-z.

Note as well how the filename was obtained by inputting a 1D character array from thekeyboard. It would seem sensible to use a string class object to accomplish this, but manycompilers do not support the use of a string object as an argument to the .open() memberfunction.

9.4.3 Reading from sequential access files

• Reading from sequential access files follows the same 6 steps except for the use of thedeclaration ifstream rather than ofstream and the direction of the data as indicatedby the symbols << for output and >> for input! That’s all there is to remember.

Here is an example called fileInput.cpp that reads the output created by the previousexample:

//File: fileInput.cpp

#include <iostream>


#include <fstream>


int main(void)

{ ifstream myInputFile; // Create an input file object

cout << "What filename do you want to read from: ";


cin >> fileName;

myInputFile.open(fileName); //Connect the stream to the file

if (myInputFile.fail())




}

else



do{

float x, y; // Two floats

myInputFile >> x >> y;;

if (myInputFile.eof()) keepReading = false; // End of file detected

else cout << x << " " << y << endl;



myInputFile.close(); //Disconnect the stream from the file

return 0;

}

Opening files safely when writing

In order to protect against destroying an existing file when opening a file for writing, youshould first open it for reading. then test to see if the file open fails. If it fails, it means thatthat file does not exist and it is safe to write to that filename. If the open does not fail, itmeans that the file exists. In this case you should prompt the user to see if the file should


be overwritten. Here’s an example:

//File: fileSafeOutput.cpp

#include <iostream>

#include <fstream>


int main(void)

{ ifstream testFile; // Create an output file object

ofstream myOutputFile; // Create an output file object

cout << "What filename do you want to write to: ";


cin >> fileName;

testFile.open(fileName);

if (!testFile.fail())

{ cout << "File called " << fileName << " already exists.\n";

cout << "Overwrite (Input 0 for no, any other int to overwrite): ";

int clobberIt;

cin >> clobberIt;

testFile.close();

if (!clobberIt) return 0;

}

myOutputFile.open(fileName); //Connect the stream to the file





}

else



do{

cout << "Two floats? ";

float x, y; // Two input floats

cin >> x >> y;;

if (cin.eof()) keepReading = false; // End of file <CNTL>-D detected

else myOutputFile << x << " " << y << endl;

9.5. COMMAND LINE ARGUMENTS 261



myOutputFile.close(); //Disconnect the stream from the file

return 0;

}

9.5 Command line arguments

Most unix commands are written in C or C++. For example, if you have ever listed thecontents of a directory, you have probably used the command ls. If you just type in ls

at the unix command prompt, you get a list of files that currently reside in your currentworking directory. If you type in ls *.cpp you will get a listing of all the files that end in.cpp. If you type ls -l, you get a “long” listing of files with information about ownershipand permissions for each file. The command is written in C and so you should wonder, “Howdid the programmer do that?”.

The programmer did it by using command line arguments. Here’s an example code thatexploits command line arguments:

//File: commandLine.cpp

#include <iostream>


int main(int argc, char* argv[])

{ cout << "argc = " << argc << endl;

cout << "This program’s name is: " << argv[0] << endl;

for (int i = 1; i < argc; i = i + 1)

{ cout << "Argument " << i << " is: " << argv[i] << endl;

}

return 0;

}

Note the new way of declaring the main routine:


{ .


.

.

}

When main is called, argc contains the number of “tokens”” (think of these as words) onthe line that ran the program. For example, if you invoked the program with

a.out word anotherWord

argc would be 3.

argv[] is an array of addresses. (The “* qualifier on a variable means that the identifiercontains one or more addresses.) These addresses contain the addresses of the starts of thetokens of the command line. Resident at these addresses is the start of the one-dimensionalcharacter arrays that contain the tokens. In the above example, the first token is the nameof the executable file (usually a.out in our examples). The second token would be word andthe third and last token in this example would be anotherWord.

here’s a program that can open a file based on a token given to it on the command line:

//File: fileOpen.cpp

#include <iostream>

#include <fstream>



{ ofstream myOutputFile; // Create an output file object

char fileName[20];

if (argc == 1)

{ cout << "What filename do you want to write to: ";

cin >> fileName;

myOutputFile.open(fileName);

}

else if (argc == 2)

{ myOutputFile.open(argv[1]);

}

else

{ cerr << "Usage: a.out [filename]\n";

}

9.5. COMMAND LINE ARGUMENTS 263



cerr << "File could not be opened.\n";


}

myOutputFile.close(); // Disconnect the stream from the file

return 0;

}

If only one token is detected, the user is prompted to input a filename. If two tokens aredetected, the second is used as a filename. If there are more than two tokens, the programstops and prints an error message.


9.6 Problems

1. Double factorials

Write a recursive function that returns the double factorial as a function of its intargument N.

• This is how the double factorial is calculated:N !! = N × (N − 2)× (N − 4) · · · (2 or 1)The algorithm stops when a reduction by 2 leaves a number that is equal to 1 or2.

• You may assume that N > 2 and you do not have to concern yourself with thefact that there is a finite (usually 32) number of bits to represent int’s.

• Your function must be compatible with the main routine below.

#include <iostream>


//Function goes here

int main(void)

{ cout << "N: ";

int N;

cin >> N;

cout << N << "!! = " << doubleFactorial(N) << endl;

return 0;

}


red% a.out

N: 5

5!! = 15

2. Sorting struct’s

This problem uses a struct defined as follows:

struct Student {int studentNumber; float finalGrade;};

9.6. PROBLEMS 265

where studentNumber is a 9-digit student number and finalGrade represents thefinal grade for that student. Write a void function that accepts one called-by-referencevector<struct Student>, and sorts the vector in descending order according to thestudent’s grade.


Chapter 10

A Potpourri of Applications

10.1 Finding a zero using the “binary chop”

Let’s say you were looking for a place where a function is zero in some interval. And let’s alsosuppose that the only thing you knew about the function was that it is monotonic (eitheralways increasing or decreasing) and that it crossed the axis only once.

In fact, you have seen such a function (in disguise) in Assignment 4. You were asked to findthe maximum of the function:

f(x) =x sin(x)

1 + x.

The derivative of this function is:

f ′(x) =sin(x)

(1 + x)2+

x cos(x)

1 + x.

f ′(x) has the properties mentioned above for certain ranges of its argument. So, finding thezero of f ′(x) is exactly the same as finding the position of the extremum (in this case themaximum) of f(x). The solution you discovered to find the maximum of f(x) involved alaborious search over the function. The algorithm about to be described, call the binarychop or mid-point bisection method, works much more efficiently, as we shall see with onlyabout 40 iterations. The pseudocode for the algorithm is:

1. Start with an initial guess for an x0 and an x1 such that the function is zero for somex0 ≤ x ≤ x1. From the properties of the function, we know that the sign of f(x0) isdifferent from the sign of f(x1).

2. Evaluate the function at the mid-point, xm = (x0 + x1)/2.

3. If f(xm) < ε , where ε is the convergence criterion, xm is the estimated zero of thefunction and the algorithm can stop. Otherwise,

267

268 CHAPTER 10. A POTPOURRI OF APPLICATIONS

4. If the sign of f(xm) is the same as the sign of f(x0), assign x1 = xm.

5. Else if the sign of f(xm) is the same as the sign of f(x1), assign x0 = xm.

6. Go back to step 2.

Here is an example of a working code.

//file findZero.cpp

#include <iostream>

#include <cstdlib>

#include <cmath>

double myF(double x)

{ static double zero = 0;

static double f = zero/zero;

static double Pi = 4*atan(1.0);

if (zero <= x && Pi >= x) f = (sin(x)/(1+x) + x*cos(x))/(1 + x);

return f;

}

int sign(double x)

{ if (0 == x) return 0;

else if (0 > x) return -1;

else if (0 < x) return 1;

else

{ cerr << "Stopping since sign can not be determined\n";

exit(EXIT_FAILURE);

}

}

int chop(double x0, double x1, double precision, double& xNew)

{ int counter = 0;

if (0 == sign(myF(x0))) // Zero at the x0 boundary

{ xNew = x0;

return counter;

}

else if (0 == sign(myF(x1))) // Zero at the x1 boundary

{ xNew = x1;

return counter;

}

while(true){ // Search for new root

counter = counter + 1;

10.1. FINDING A ZERO USING THE “BINARY CHOP” 269

xNew = (x0 + x1)/2;

if (precision > fabs(myF(xNew))) return counter;

if (sign(myF(xNew)) == sign(myF(x0))) x0 = xNew;

else x1 = xNew;

}

}

bool badLimits(double x0, double x1)

{ if (sign(myF(x0)) == sign(myF(x1))) return true;

return false;

}

int main(void)

{ double x0, x1, x;

do

{ cout << "Two limits? ";

cin >> x0 >> x1;

}while(badLimits(x0,x1));

double precision;

cout << "Precision? ";

cin >> precision;

cout << "\n\nFinding root for function between "

<< x0 << " and " << x1 << " to precision = "

<< precision << endl;

cout << "Number of iterations = "

<< chop(x0,x1,precision,x) << endl;

cout << "Function is near 0 at x = " << x << endl;

cout << "Value of function there = " << myF(x) << endl;

return 0;

}

Apart from the idea of the algorithm, there are a few new features in this code. One is theoutput stream cerr. cerr is a different output stream than cout and it is intended to be thestream to which diagnostic or error messages are sent. For example, the "int sign() functionabove writes to cerr when it can not determine the sign of the function.

The other important use of cerr is when the user of a program has redirected cout to a filevia the unix redirection symbols. For example, output to cout could be captured in a filenames myOutput as follows:red% a.out > myOutput

270 CHAPTER 10. A POTPOURRI OF APPLICATIONS

However, cerr would still print to the screen so that the user would know something hadgone wrong. If the error message had been written to cout, the user would not know untilhe or she opened to file.

Another example is:red% a.out < myInput > myOutput

which reads cin from the file called myInput rather than the keyboard. Note that the filenames are arbitrary. The names myOutput and myInput are just examples.

Chapter 11

Programming in MATLAB

11.1 A taste of MATLAB

• Chapter 11: Basic info, logical operators, logic flow control, functions (read now)

• Chapter 12: 2D and 3D graphics

• Chapter 13: Some applications

What is MATLAB?

• An interpreted computer language

• The name of the program that interprets MATLAB language

• Central idea—all data is represented by 2D arrays of doubles

Data types and variables:

• Names of variables? Same as ANSI C++

• Types of variables and declarations?

C++: Many types, MUST declare

MATLAB: 1 type (2-D arrays of doubles) only and don’t declare

• A MATLAB demo:

271

272 CHAPTER 11. PROGRAMMING IN MATLAB

>> i = 10;

>> disp(i)

10

>>

• Statements

ANSI C++: Terminated by a semicolon

MATLAB: Terminated by and end-of-line (carriage return)unless you put a ... at the end of the line to continueCan also terminate with a ; which suppresses the echo of the statement

• Flow control—while loops

ANSI C++:

i = 0;

while(i<10)

{ i = i + 1;

}

MATLAB:

i = 0;

while(i<10)

i = i + 1;

end

• Flow control—if/else constructs

ANSI C++:

if (i <= 10)

{ j = 0;

}

else

{ k = 0;

}

MATLAB:

11.2. ARRAYS IN MATLAB 273

if (i <= 10)

j = 0;

else

k = 0;

end

• Example of summing the integers, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

ANSI C++:

int main(void)

{ int i = 0, s = 0;

while(i < 10)

{ i = i + 1;

s = s + i;

}

cout << s << "\n");

return 0;

}

MATLAB:

i = 0;

s = 0;

while(i < 10)

i = i + 1;

s = s + i;

end

disp(s)

• Scripts and functions

• Scripts (or script M-files) are MATLAB statements in a file that the MATLAB in-terpreter will execute

Functions (or function M-files) are used to implement functions.Function names are derived from file names.

11.2 Arrays in Matlab

• The idea: 2-D arrays of doubles are the fundamental data type


• Creating arrays in MATLAB

– zeros —Creates and initializes arrays in MATLAB and fills the array with “0.0”’s

– ones —Creates and initializes arrays in MATLAB and fills the array with “1.0”’s

– Some examples:

>> x = ones(10,1)

x =

1

1

1

1

1

1

1

1

1

1

>> whos

Name Size Bytes Class

x 10x1 80 double array

Grand total is 10 elements using 80 bytes

>> z = zeros(5,3)

z =

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

>> whos



z 5x3 120 double array



>>

– The colon operator (:) gives a list of “integers”Example:

>> clear % This is how to clear stored variables

>> % This is how to make a comment

>> x = 1:10

x =

1 2 3 4 5 6 7 8 9 10

>> whos




>>

So x = 1:N creates a row with values 1, 2, 3...N

– You can use two colon operators (a:∆:b) to make a list of the form:a, a+∆, a+ 2∆, a+ 3∆ · · ·a+N∆ where a+N∆ ≤ b < a+ (N + 1)∆.

>> y = 1:2:10

y =

1 3 5 7 9

Note that the upper limit, 10, is not reached because the next number in theseries would overrun it.

– linspace —creates a 1D array with uniformly spaced elements.x = linspace(a,b,N)—makes N equally spaced points between a and b inclusive.

(The statement x = a:(b-a)/(N-1):b seems to accomplish the same thing BUTround-off errors may cause the end-point b to be excluded! DO NOT USE the x= a:(b-a)/(N-1):b form!)

An example:


>> clear

>> x = linspace(0,1,5)

x =

0 0.2500 0.5000 0.7500 1.0000

>>

– Literal creation:

∗ Create an array with one row and 3 columns:

>> x = [1.0,5.2,9.7]

x =

1.0000 5.2000 9.7000

>>

∗ Create an array with 2 rows and 3 columns:

>> y = [1.0,5.2,9.7;3.4,9.3,10.7]

y =

1.0000 5.2000 9.7000

3.4000 9.3000 10.7000

>>

• Array indexing

– Simple indexing

x(i) refers to the i’th element of array x

There is no off-by-one nonsense!

x = ones(1,10) =⇒ x = [1,1,1,1,1,1,1,1,1,1]

x(1) is the first element of array x

x(10) is the last element of array x

Array indices go from 1 to the number of rows/columns

>> x = [3,5;7,9]

x =

3 5

7 9


x(1,1) = 3 =⇒ row 1, column 1x(2,1) = 7 =⇒ row 2, column 1x(1,2) = 5 =⇒ row 1, column 2x(2,2) = 9 =⇒ row 2, column 2

– Slice indexing

The “:” symbol as an array index means “the whole” row or “the whole” column.Example,

>> x = [1,2;3,4;5,6]

x =

1 2

3 4

5 6

>> y = x(:,2)

y =

2

4

6

>> z = x(1,:)

z =

1 2

Partial slices can be extracted by specifying ranges. Example,

>> u = x(1:2,1)

u =

1

3

>> v = x(2:3,:)

v =


3 4

5 6

• Saving an array to a file

sindat is a 2 row, 1000 column array created and saved this way:

%ch11e1.m

sindat = zeros(2,1000);

sindat(1,:) = linspace(0,2*pi,1000);

sindat(2,:) = sin(sindat(1,:)); % Note: array operation!

save sindat -ascii -double

• Reading an array from a file

The following reads sindat in and plots it.

%ch11e2.m

clear

load sindat

x = sindat(1,:);

y = sindat(2,:);

plot(x,y) % Simple plotting function

pause % Pauses until any keystroke is given

close % Closes the figure

plot(x,y.^2) % Note: ".^2" means element by element squaring operation

pause

close

plot(y,x)

pause

close

Summary

• zeros—Create and initialize an array and fill the array with “0.0”’s

• ones—Create and initialize an array and fill the array with “1.0”’s

• clear—Clear variables from memory

• whos—List current variables, long form

• linspace—Create a 1D array with uniformly spaced elements

11.3. LOOPS 279

• Literal creation of arrays

• Simple array indexing

• Slice indexing, full and partial slices

• Writing/reading arrays to/from files.

11.3 Loops

11.3.1 The for loop

The general syntax of a for loop is:

for i = array

% Some loop using i where i can not be re-assigned

end

On each successive pass through the loop, i is set to the next column of array. On the firstpass through the loop, i is array(:,1), on the second pass i is array(:,2), etc

This can seem a little confusing unless we remember that the fundamental data type is anarray of doubles.

Here are some simple examples:

sum = 0;

for i = 1:10

sum = sum + i;

end

disp(sum);

Here’s another way of doing the same thing:

i = 1:10;

sum = 0;

for j = i

sum = sum + j;

end

disp(sum);

Here’s yet another way of doing the same thing:


i = 1:10;

sum = 0;

for j = 1:length(i) % length(array) is a MATLAB function that gives the length

sum = sum + j;

end

disp(sum);

Here’s a weird way of doing the same thing:

i = [1,2,3,4,5;6,7,8,9,10];

sum = 0;

for j = i % j is set to the columns of i

sum = sum + j(1) + j(2);

end

disp(sum);

Why does this work?The number of columns of i is 5 and so, the loop is executed 5 times. But on each iteration,y is set to a column of x, in this case a column vector with two values which are summedwith the statement sum = sum + j(1) + j(2);. Weird! But it works and illustrates themore general form of the looping index.

Some useful functions

size gives the number of rows and the number of columns as a two-element row vector.For example (using the i array above):

>> size(i)

ans =

2 5

If you want to “capture” both the number of rows and columns, use the following:

>> [nRows,nColumns]=size(i)

nRows =

2

nColumns =

5

length gives the maximum of the number of rows and the number of columns. For example

(using the i array above):

11.3. LOOPS 281

>> length(i)

ans =

5

error error(’string’) displays the ’string’ and causes an error exit from an M-file to thekeyboard. If the string is empty, no action is taken.

Here’s an example of an integration script using length and error.

% Area under f as a MATLAB script

% Assumes that f and x already exist

if (length(x) ~= length(f))

error(’ x and f are different lengths. Stopping.’)

end

sum = 0;

for j = 1:(length(f)-1)

sum = sum + 0.5*(f(j) + f(j + 1))*(x(j+1) - x(j));

end

disp(sum);

Here’s a 2D plotting example:

N = 100;

f = zeros(1,N);

x = linspace(-5*pi,5*pi,N);

for i = 1:N

if (x(i) == 0)

f(i) = 1;

else

f(i) = sin(x(i))/x(i);

end

end

plot(x,f);

Here’s a 3D plotting example (using the above f):

g = zeros(N,N);

for i = 1:N

for j = 1:N

g(i,j) = f(i)*f(j);


end

end

surf(x,x,g);

11.3.2 The while loop

The general syntax of a while loop is:

while expression

% Some loop that is executed when expression is true

end

Here’s an example

change = 1;

while change >= 0.001

change = change/2;

end

disp(change);

That’s about all you need to know about while loops.

11.4 The if/else construct

The general syntax of an if/else if/else construct is:

if expression1

% Executed when expression1 is true

elseif expression2

% Executed when expression1 is false and expression 2 is true

elseif expression3

% Executed when expression1&2 are false and expression 3 is true

.

.

.

elseif expressionN

% Executed when expression1...N-1 are false and expression N is true

else

11.5. SOME MATLAB TERMINOLOGY 283

% Executed when expression1...N are false

end

Valid syntax requires the if and the end, zero or more elseif’s and zero or one else. Thedesign is very much like C and therefore does not really need to be discussed further. Theuse of indenting make the logical flow a little more comprehensible.

Here’s an example that makes a decision based on a random number:

x = rand;

if x < 1/3

disp(’x is less than a third’);

elseif x < 2/3

disp(’x is greater than or equal to a third but less than two thirds’);

else

disp(’x is greater than or equal to a two thirds’);

end

New function:rand —provides an array of uniformly distributed random numbers between 0 and 1.

11.5 Some Matlab terminology

scalar A scalar in MATLAB is a 1× 1 array. For example:

>> whos


s 1x1 8 double array


row vector A row vector of length M in MATLAB is a 1×M array. For example:

>> rv = ones(1,8);

>> whos


rv 1x8 64 double array



column vector A column vector of length N in MATLAB is an N × 1 array. For example:


>> cv = zeros(10,1);

>> whos


cv 10x1 80 double array




matrix An N by M matrix in MATLAB is an N ×M array. For example:

>> a = rand(3,4);

>> whos


a 3x4 96 double array

cv 10x1 80 double array




>> a

a =

0.7382 0.9355 0.8936 0.8132

0.1763 0.9169 0.0579 0.0099

0.4057 0.4103 0.3529 0.1389

11.6 Operators in MATLAB

11.6.1 Math operators in Matlab

Operation Symbol Example PrecedenceInner expression (,) (1+2)/3 HighestExponentiation ^ 2^3 Medium highMultiplication, x× y * x*y ModerateDivision, x/y, y\x /,\ y/x = x\y ModerateAddition, x+ y + x + y Medium lowSubtraction, x− y - x - y Medium lowAssignment = y = x Lowest

These are all 2-D matrix operations!If the operands are scalars (a 1× 1 array) the functionality is similar to that of C++.

Here are some examples:

11.7. POINTWISE OPERATORS IN MATLAB 285

>> a = 2*ones(2)

a =

2 2

2 2

>> b = a/4

b =

0.5000 0.5000

0.5000 0.5000

>> a*b

ans =

2 2

2 2

>> a+b

ans =

2.5000 2.5000

2.5000 2.5000

11.7 Pointwise operators in MATLAB

Operation Symbol Example ExpansionPointwise exponentiation .^ x.^y x(i,j)^y(i,j)

Pointwise multiplication .* x.*y x(i,j)*y(i,j)

Pointwise division ./,.\ y./x = x.\y x(i,j)/y(i,j)

Here are some examples:

>> x = 2*ones(2)

x =

2 2

2 2

>> y = [0,1;2,3]

y =

0 1

2 3

>> x.*y

ans =

0 2

4 6

>> x.^y


ans =

1 2

4 8

>> x.\y

ans =

0 0.5000

1.0000 1.5000

11.7.1 Logical Operators in Matlab

Symbol Relational operation< , <= less than, less than or equal to> , >= greater than, greater than or equal to== equivalent to~= not equivalent to& , | AND , OR~ NOT

Note that these operators work on arrays and can return arrays. The relational operatorsoperating on arrays is not discussed in this course; we will be using them as if they alwaysoperate on scalars. Therefore, in this sense, they behave exactly as the analogous operatorsin C++,

There are some differences in notation with C++ (~= vs. !=, & vs. &&, | vs. ||, ~ vs.!) However, the biggest difference with respect to C++ is the difference in the order ofprecedence of the operators. Note that even some textbooks are confused on this issue. Hereis the current state of the order of precedence of the MATLAB operators. Note that forMATLAB Version 5.3 and lower, the precedence table is different!

Operation Symbol PrecedenceInner expression ( ) 0 (Highest)Transpose, Exponentiation ^ .^ ’ 1Unary plus and negation + - ~ 2Multiplication, division .* .\ ./ * \ / 3Binary addition, subtraction + - 4Colon operator : 5Relational logical operators < <= > >= == ~= 6Logical AND & 7Logical OR | 8Assignment = 9 (Lowest)

For more complete information, issue help precedence. This is what you get:

PRECEDENCE Operator Precedence in MATLAB

11.7. POINTWISE OPERATORS IN MATLAB 287

MATLAB has the following precedence for the built-in operators when

evaluating expressions (from highest to lowest):

1. transpose (.’), power (.^), complex conjugate

transpose (’), matrix power (^)

2. unary plus (+), unary minus (-), logical negation (~)

3. multiplication (.*), right division (./), left

division (.\), matrix multiplication (*), matrix right

division (/), matrix left division (\)

4. addition (+), subtraction (-)

5. colon operator (:)

6. less than (<), less than or equal to (<=), greater than

(>), greater than or equal to (>=), equal to (==), not

equal to (~=)

7. logical AND (&)

8. logical OR (|)

Starting in MATLAB 6.0, the precedence of the logical AND (&)

and logical OR (|) operators now obeys the standard relationship

(AND being higher precedence than OR) and the formal rules

of boolean algebra as implemented in most other programming

languages, as well as Simulink and Stateflow.

Previously MATLAB would incorrectly treat the expression:

y = a&b | c&d

as:

y = (((a&b) |c) &d);

It now correctly treats it as:

y = (a&b) | (c&d);


The only case where the new precedence will impact the result

obtained is when | appears before & within the same expression

(without parentheses). For example:

y = 1 | x & 0;

In MATLAB 5.3 and earlier, this statement would yield 0,

being evaluated as:

y = (1 | x) & 0;

In MATLAB 6.0 and beyond, this expression yields a 1 as the result,

being evaluated as:

y = 1 | (x & 0);

We strongly recommend that you add parentheses to all expressions of

this form to avoid any potential problems with the interpretation

of your code between different versions of MATLAB.

NOTE: A feature has been provided to allow the system to produce an

error for any expression that would change behavior from the old to

the new interpretation:

feature(’OrAndError’, 0) % warning for any expression that

% changed behavior (default)

feature(’OrAndError’, 1) % error for any expression that

% changed behavior

11.8 M-files

11.8.1 Script M-files

MATLAB commands may be stored in a file and executed. These files are called “scriptM-files”. Any commands that you can enter in the MATLAB command window can beentered in a file with a filename that is arbitrary except that it should contain the extension.m, for example myMatlabScript.m. Script M-files may be created with your favorite editor

11.8. M-FILES 289

or with MATLAB’s own editor.

If you type the name, say myMatlabScript in the MATLAB command window, the followingorder of execution is followed:

• MATLAB checks to see if myMatlabScript is a variable in the current workspace, theenvironment of user-defined and pre-defined variables; if not

• MATLAB checks to see if myMatlabScript is built-in function; if not

• MATLAB checks to see if myMatlabScript.m is a file in the current directory; if not

• MATLAB checks to see if myMatlabScript.m exists anywhere along the MATLABsearch path;

Some useful functions for use with MATLAB script M-files:

disp(variable) displays the value of the variable, e.g..

>> n = 10; disp(n)

10

>>

echo toggles echoing.echo off turns it off (default value)echo on turns it on and every command executed by a script M-file is written to the screen(useful for debugging purposes)echo switches (toggles ) between echo off and echo on.

input prompt the user for input.

>> x = input(’Input a value for x: ’)

Input a value for x: 10

x =

10

>>

keyboard gives control to the keyboard. Return to the script M-file by typing return at

the command window prompt, (>>).

pause Pause. Continue after user presses any keyboard key.

pause(x) Pause for x seconds, then continue.


>> for i = 10:-1:1

disp(i)

pause(1)

end

10

9

8

7

6

5

4

3

2

1

>>

waitforbuttonpress Pause. Continue after user presses any keyboard key or mouse key.

sprintf Write formatted data to string.S = sprintf(FORMAT,A,...)

formats the data in the variable A (and in any additional arguments), under control of thespecified FORMAT string, and returns it in the MATLAB string variable S. There is a similarfunction called fprintf that writes the MATLAB string variable to a file. (Do a help

fprintf for more information.)

FORMAT is a string containing C++ language conversion specifications. Conversion speci-fications involve the character %, optional flags, optional width and precision fields, optionalsubtype specifier, and conversion characters d, i, o, u, x, X, f, e, E, g, G, c, and s. Seethe Language Reference Guide or a C++ manual for complete details. The %g conversionspecifier produces the nicest output for floating point numbers. See the example below.

The special formats \n,\r,\t,\b,\f can be used to produce linefeed, carriage return, tab,backspace, and formfeed characters respectively. Use \\ to produce a backslash characterand %% to produce the percent character.

Here’s an example of a script M-file called russianRoulette.m:

%File: russianRoulette.m

clear

echo off

prob = 1/6;

disp(’With each spin you could win 1$’)

11.8. M-FILES 291

disp(’If you lose, you lose everything!’)

disp(’It’’s best to quit while you’’re ahead.’)

disp(’Press any key to start.’)

disp(’Good luck’)

disp(’ ’)

pause

money = 0;

repeat = input(’Play? (0 = no, 1 = yes) :’);

while(repeat)

disp(’Spin....’)

pause(1)

if (rand < prob)

disp(’BANG! You LOSE!’)

return

else

disp(’Click! You WIN!’)

money = money + 1;

disp(sprintf(’Your gain is $%g’,money))

end

repeat = input(’Repeat? (0 = no, 1 = yes) :’);

end

disp(sprintf(’Your gain was $%g’,money))

disp(’End of Game’)

disp(’ ’)

Here’s a typical play of the game:

>> russianRoulette

With each spin you could win $1.

If you lose, you lose everything!

It’s best to quit while you’re ahead.

Press any key or mouse button to start.

Good luck!

Play? (0 = no, 1 = yes) :1

Spin....


Click! You WIN!

Your gain is $1

Repeat? (0 = no, 1 = yes) :1

Spin....

Click! You WIN!

Your gain is $2

Repeat? (0 = no, 1 = yes) :1

Spin....

Click! You WIN!

Your gain is $3

Repeat? (0 = no, 1 = yes) :1

Spin....

Click! You WIN!

Your gain is $4

Repeat? (0 = no, 1 = yes) :1

Spin....

Click! You WIN!

Your gain is $5

Repeat? (0 = no, 1 = yes) :1

Spin....

Click! You WIN!

Your gain is $6

Repeat? (0 = no, 1 = yes) :1

Spin....

Click! You WIN!

Your gain is $7

Repeat? (0 = no, 1 = yes) :1

Spin....

BANG! You LOSE!

11.8.2 Function M-files

Compartmentalizing your code into functional subunits is...yada, yada, yada...It’s the same sermon you got in C++! However, MATLAB makes it easy to designfunctions for repeated use in many applications.

The general syntax of a function call is:

[out1,out2,...,outN] = functionName(in1,in2,inM)

and the general syntax of the function definition is to have a separate file called functionName.mthat has the following form:

11.8. M-FILES 293

function [out1,out2,...,outN] = functionName(in1,in2,...inM)

%FUNCTIONNAME A brief one line description

% A more general description of functionName with a description of the inputs and

% outputs and perhaps an example. All this text must fit into contiguous comment

% lines.

%

% See also related list of functions

% After this line starts MATLAB executable functions, important examples below

.

.

.

error(’This is an error’) % Can use this to return control to the command

% windows if an error is detected

.

.

.

warning(’This is a warning’) % Can use this to issue warnings

.

.

.

return; % Control reverts to calling M-file if a return is seen

.

.

.

% Last line in functionName, control reverts to calling M-file

Some general features of functions:

• functionName is an identifier, identical to that of variables

• Variables can only be output in the output “string” [out1,out2,...,outN]. Therecan be zero or one output (in which case the square brackets are not needed).

• There can be zero, one or more inputs. If you change the value of an input variable itis hidden from the calling M-file. If you do not change the value of an input, MATLABwill not make copies for the function’s workspace for the sake of efficiency. If you dochange the value of an input, MATLAB will make copies for the function’s workspaceand the changes will be hidden from the calling M-file.

• The command window and the functions have their own workspaces, places wherevariables are stored independently, much like C++’s stack frame handling.

• The first line of the function M-file is called the function declaration line. The functionname must agree with the file name. If it does not, MATLAB will use the file name.


This line also serves a similar role as the function prototype in C++, checking the com-patibility of the function definition with the function call. (MATLAB is considerablylooser, however, than ANSI C++).

• If the second line of the file is a comment, it is used by MATLAB to look for functionsand functionality using the lookfor MATLAB function.

• Comment lines following contiguously are copied to the screen when you typehelp functionName

• You can have more than one function in a function M-file. However, only the first (theprimary function) is available to other M-files and the command window. The othersare called subfunctions and cannot be used by other functions, scripts or the commandwindow.

Here is an example called quadRoots.m:

function [root1,root2] = quadRoots(A,B,C)

%QUADROOTS quadRoots returns the two real roots of A*x^2 + B*x + C = 0

%

% The two roots are root1 = (-B + sqrt(B^2 -4*A*C))/(2*A)

% root2 = (-B - sqrt(B^2 -4*A*C))/(2*A)

%

% An error is detected if 4*A*C > B^2 (imaginary roots)

% A warning is issued if 4*A*C = B^2 (identical roots)

%

% The blame? BLIF <[email protected]>

% This version 1.1, last modified November 27, 2000

% Warranty? None. Take it or leave it.

% Copyright: Regents of the University of Michigan

if 4*A*C > B^2

error(’Roots would be imaginary. Get real!’)

elseif 4*A*C == B^2

warning(’Roots will be the same, degenerate’)

end

root1 = (-B + sqrt(B^2 -4*A*C))/(2*A);

root2 = (-B - sqrt(B^2 -4*A*C))/(2*A);

return % This is optional

11.9. PROBLEMS 295

11.9 Problems

1. Check the appropriate circle for the following MATLAB statements.

MATLAB Statement Valid Invalidy=3^2; © ©if(23==x) © ©23=x; © ©for i=-123:234 © ©while( ~(~(~(x<3))) ) © ©a(1:10)=1; © ©b=[1 2 3; 4 5;6 7]; © ©

2. Consider the following snippet of code

for p = 1:10

for g = 1:10

x(g,p) = g*p;

end

end

This piece of code:

© Fills the entire first row of the array, then moves on to the next row, and so on tofill the array.

© Fills the entire first column of the array, then moves on the next column, and soon to fill the array.

3. For each of the following tasks, check the appropriate box to state which construct isthe most natural way of implementing that task.

Counter loop Sentinel loop if/elseInitializing the elements of an array ofknown dimensions

© © ©

Responding to a user’s selection from amenu

© © ©

Carrying out an iterative process untilsome measure of convergence is reached

© © ©

Dealing a hand of cards from a deck © © ©Determining a student’s letter grade forthis final from his or her numerical score

© © ©

Computing the average score on this finalusing the scores of a known number of stu-dents

© © ©


4. Consider the following small script M-file and function M-file.

x = 4;

y = [2 3;6 7];

a(1,1) = dumbFunction(x,y(2,1));

function r = dumbFunction(a,b)

r = b/a;

return;

Now, complete the following sentence by filling in the blanks.

The numerical values and are passed into the functiondumbFunction. The function returns the numerical value , whichis then stored in the variable .


List and describe the three capabilities that are needed to construct an algorithm.

6. MATLAB Function M-file design

Describe the roles of:

(a) the first line of a MATLAB function M-file,

(b) the second line of a MATLAB function M-file (assuming it starts with a “%”),

(c) the third and subsequent contiguous comment lines in a MATLAB function M-file(assuming the second line starts with a “%”).

How do the MATLAB lookfor and help functions interact with these lines?

7. MATLAB Function M-files

Write a MATLAB Function M-file that provides the indicated response to the lookforand help commands. A human typed the stuff to the right of a >> prompt, thecomputer typed the rest.

>> lookfor helpMe

HELPME Help! I need somebody.

>> help helpMe

HELPME Help! I need somebody.

Help! Not just anybody!

>>

11.9. PROBLEMS 297

8. True or false?

(a) The 3 statements

clear all; x = 1.0; i = &x;

are ALL valid MATLAB statements.

(b) The 3 statements

clear all; x = 1; i = i & x;

are ALL valid MATLAB statements.

(c) The set of statements

clear all; a = 2*ones(2,2); b = a/4; c = a*b; disp(c)

prints 1 1 to the screen.1 1

(d) The set of statements (note use of the transpose operator ”’”)

clear all; disp(linspace(0,1,5)’)

prints 0 0.2500 0.5000 0.7500 1.0000 to the screen.

(e) The statement

a = [0,1,2; 2,3; 4,5;];

is a valid MATLAB statement.

9. Multiple choice

Circle all the correct answer(s). There may more than one correct answer but therewill be at least one choice that is correct.

(a) The following MATLAB code:

clear all; N = 5; a = zeros(N);

for i = (2:1:N)

a(i,i-1) = -1;

if i ~= N a(i,i+1) = 1; end

end

disp(a)

is supposed to print to the screen:


0 1 0 0 0

-1 0 1 0 0

0 -1 0 1 0

0 0 -1 0 1

0 0 0 -1 0

but it does not. The error is

• There is an syntax error in the line: if i ~= N a(i,i+1) = 1; end

• After a = zeros(5);, nothing is changed in the first row.

• There is an syntax error in the line: for i = (2:1:N)

• There is a conceptual error in the statement: a(i,i-1) = -1;

• None of the above, there is a syntax error somewhere else.

(b) The following code is supposed to sum the squares of the matrix elements of arandom 10 by 10 array and then divide the sum by the number of elements. Itdoes not work as intended.

clear all;

N = 10;

s = rand(N);

s2 = s*s;

for i = 1:N for j = 1:N sum = sum + s2(i,j); end; end

disp(sum/N^2)

• N is not defined properly.

• There is a syntax error in the line that starts with for.

• sum should be initialized first, before the for loops.

• There is a syntax error in the statement disp(sum/N^2).

• There is a conceptual error in the statement s2 = s*s;.

• There is no error. The code works as intended.

• There is a conceptual error in the statement s = rand(N);.

10. Slice indexing

The following code is written with 4 for loops.

N = 100;

a = zeros(N);

for i = 1:N

for j = 1:N

a(i,j) = 1;

end

end

11.9. PROBLEMS 299

for i = N/4:3*N/4

for j = N/4:3*N/4

a(i,j) = 0;

end

end

(a) Write it without the use of for or while loops.

(b) If the variable N was set to 4 instead of 100, what would the array look like?

11. Think like a computer

Write the output that the following MATLAB code produces.

clear all;

z = ones(6);

z(3:4,:) = 0;

z(:,3:4) = 0;

disp(z);

12. Think like a computer again

Consider the following script M-file.

sign = 1;

for i = 1:5

x(i) = i - 3;

y(i) = sign*i;

sign = -sign;

end

plot(x,y);

Draw the plot that would result from executing this script.


Write the output that the following MATLAB code produces.

clear all;

z = -ones(7); % Don’t miss the minus sign!

z(2:3,2:3) = 1; z(2:3,5:6) = 4;

z(5:6,2:3) = 2; z(5:6,5:6) = 3;

z(4,4) = 0;

disp(z);



Consider the following script M-file.

a = [1 2];

b = [1 2 3];

c = 1;

for ai = a

for bi = b

fprintf(’a = %f, b = %f\n’,ai,bi)

if (bi^2 - 4*ai.*c < 0)

fprintf(’No real solution\n\n’)

elseif (bi^2 - 4*ai.*c == 0)

fprintf(’Single root = %f\n\n’,-bi/(2*ai))

else

fprintf(’Root1 = %f\n’,(-bi - sqrt(bi^2 - 4*ai.*c))/(2*ai))

fprintf(’Root2 = %f\n\n’,(-bi + sqrt(bi^2 - 4*ai.*c))/(2*ai))

end

end

end

What is the output that would result from executing this script?

15. The purpose of the script M-file below is to:

• prompt the user for a number between 0 and 100

• generate a row-vector of 50 random numbers between 0 and 100

• invoke a function called nearest

• print out the results returned from the function

The purpose of the function M-file below is to :

• take in a row-vector and a scalar value

• find the entry in the row-vector that is closest to the scalar value

• return the entry that was found, and its index in the array

There are four bugs in the code, at lines 5, 7, 17 and 18. Find them and fix them.

% Script M-file testNearest.m

%

x = input(’pick a number between 0 and 100: ’); 1

11.9. PROBLEMS 301

2

b=rand(1,50)*100; 3

4

[ind,y] = nearest(b(50),x); 5

6

fprintf(’The nearest value to %f is b(%i)=%f\n’,x,ind,b); 7

% Function M-file nearest.m

function [ind,y] = nearest(a,x) 8

9

oldDiff = 1000; 10

[rows,columns] = size(a); 11

12

for i = 1:columns 13

newDiff = abs(a(i) - x); 14

if(newDiff < oldDiff) 15

y = a(i); 16

index = i; 17

newDiff = oldDiff; 18

end 19

end 20

21

return 22

16. Plot like a computer

The code below produces a 2D plot. Indicate what it would look like in the box below.

clear all;

x = [0,-0.59, 0.95,-0.95, 0.59, 0];

y = [1,-0.81, 0.31, 0.31, -0.81, 1];

plot(x,y)

17. More code debugging

In the following code I want to plot f(x) = sin(x)x

for 0 ≤ x ≤ 5π by making two rowvectors, one for x and the other for f(x). The code does not generate any MATLABerrors and produces a plot. However, the plot does not look anything like what Iexpected. Indicate what needs to be done to fix it by writing a correct version of thecode in the box below. The code below has conceptual problems on lines 3 and 5. Takecare that your code does not have a “divide-by-zero” problem. (Hint: f(x) = 1 whenx = 0.)


clear all; % Line 001

N = 1000; % Array size Line 002

f = zeros(1000); % Allocate memory for f Line 003

x = linspace(0,5*pi,1000); % Make a linspace for x Line 004

f = sin(x)/x; % Calculate f(x) Line 005

plot(x,f) % Plot it Line 006

18. Temperature conversion

Write a code that outputs a table of degrees Fahrenheit and degrees Centigrade startingwith -40 degrees Fahrenheit, ending with 140 degrees Fahrenheit in increments of 1degree Fahrenheit. The conversion from degrees Fahrenheit to degrees Centigrade is:C = (5/9) ∗ (F − 32)and must be coded as a separate MATLAB function. The output should look like:

-40,-40.0

-39,-39.4

-38,-38.9

.

.

.

138,58.9

139,59.4

140,60.0

Hint: The format statement that outputs one line of the table isfprintf(’%3.0f,%4.1f\n’,F(i),C(i)).

19. Array manipulation

Create a complete set of MATLAB instructions that makes a 1000 by 1000 matrixcalled “a” that is zero everywhere except for ones in the first column and last columnand ones in the first row and last row. There are also ones on the main diagonal of thematrix, (a(i,i) = 1) for all i. For full credit you can not use more than one for orwhile loop. Otherwise, one point will be deducted for each use of a for or a while

loop. Do it the MATLAB way! There are five extra credit points on this problem forthose who remember and use correctly (from reading the book, or from attending thelast lecture) the command for making an array with ones on the main diagonal andzeros everywhere else.

20. Matrix rotation and flipping

Write 4 different functions that:

• Rotates a matrix in a clockwise fashion

11.9. PROBLEMS 303

• Rotates a matrix in a counterclockwise fashion

• Flips a matrix about its horizontal mid-line

• Flips a matrix about its vertical mid-line

Hints and instructions.

• Recall that [Rows,Columns] = size(a) can be used to get the number of rowsand columns of the matrix a

• Your functions have to work on arrays of any size

• Your functions do not have to provide output to the help and lookfor functions.

Here’s an example:

>> a = rand(2,3)

a =

0.9218 0.1763 0.9355

0.7382 0.4057 0.9169

>> disp(rotateClockWise(a))

0.7382 0.9218

0.4057 0.1763

0.9169 0.9355

>> disp(rotateCounterClockWise(a))

0.9355 0.9169

0.1763 0.4057

0.9218 0.7382

>> disp(flipVertical(a))

0.9355 0.1763 0.9218

0.9169 0.4057 0.7382

>> disp(flipHorizontal(a))

0.7382 0.4057 0.9169

0.9218 0.1763 0.9355

21. Stars and Stripes forever

At the end of this problem description, parts of a script M-file are given. This programmay be used to draw the US flag. The US flag has 13 horizontal rows, alternating withred and white. Row 1 is red, row 2 is white, and so on, for 7 red rows and 6 whiteones. A solid blue rectangle occupies the upper left corner, from rows 7 to 13 and it ishalf the width of the flag. There are 50 white stars arranged in the blue patch.

Here’s how you complete the code to finish the flag program.

• Put in the code that will make the stripes and the blue patch. (Red value 0.15,blue has value 0.4, and white has value 0.0.) Plotted as indicated it should looklike:


• Read through the rest of the MATLAB code and understand what it is trying todo. The bottom part is for placing the stars.

• Write the function star(x,y,r). x and y are the points that define the center ofthe star, and r is its radius. Notice that the 5 points on a star are all located on acircle centered at x and y with radius r. If you drew lines from the center to thepoints of the star, those lines would be separated by 72 degrees, (72 = 360/5).To draw the star, imagine that the top point of the star is numbered 1, the oneto its right 2, and so on. You can make a star easily by drawing a line from point1, to 4, to 2, to 5, to 3 and back to 1. Here’s an example of star(1,3,2).

• You might find it useful to to use the sin() and cos() functions. The argumentto these functions should be expressed in radians. In MATLAB, the conversionfrom degrees to radians is given by: rads = pi*a/180, where a is an angle indegrees and rads is the same angle expressed in radians.

• You should use the plot() function using white solid lines. Here are the details:

PLOT Linear plot. PLOT(X,Y) plots vector Y versus vector X.

Various line types, plot symbols and colors may be obtained with

PLOT(X,Y,S) where S is a character string made from one element

from any or all the following 3 columns:

b blue . point - solid

11.9. PROBLEMS 305

−1 −0.5 0 0.5 1 1.5 2 2.5 31

1.5

2

2.5

3

3.5

4

4.5

5

g green o circle : dotted

r red x x-mark -. dashdot

c cyan + plus -- dashed

.

.

.

For example, PLOT(X,Y,’c+:’) plots a cyan dotted line with a plus

• Your function has to respond sensibly to the lookfor and help functions.

%FLAG flag Solution to Stars and Stripes forever

clear all

%Code to make the flag without stars below


%Show the flag without stars using pcolor

show = flag;

show(end+1,1) = 0; %Hidden point, fixes the lower bound of the colormap

show(end,end+1) = 1; %Hidden point, fixes the upper bound of the colormap

pcolor(1:41,1:14,show)

colormap vga

shading flat

axis off

hold on

%This code prompts the user to place the stars...

putStar = 1;

while putStar

putStar = input(’Place a star? (0 = no, ~0 = yes)’);

if putStar

response = input(’Specify [x, y, size] ’);

x = response(1); y = response(2); r = response(3);

star(x,y,r)

end

end

22. Turn pseudo-code into real code

Write a MATLAB code to find the zero of a function. Your code will be made up ofthree M-files:

• a script M-file to drive the process

• a function M-file that finds zeros using a Newton procedure (explained below)

• a function M-file that contains the function for which you are finding the zero.

Pseudo-code for the three pieces is given below. Construct the MATLAB code.

Script M-file

(a) Prompt the user for an initial guess for the x value at which the zero occurs.

(b) Call the zero-finding function, with the initial guess as the input argument,and the final answer and number of steps it took to find the final answer asthe output arguments.

(c) Print out the final answer and the number of steps it took to find the finalanswer in some self-explanatory format.

Function M-file for Finding a Zero

11.9. PROBLEMS 307

(a) This function has one input argument, the initial guess for x.

(b) This function has two output arguments, the final approximation of the x-value at which the zero is found, and the number of steps it took to findit.

(c) Set a convergence criterion of ε = 10−6.

(d) Initialize the number of iterations to zero.

(e) Carry out the following steps until the absolute value of f(x) < ε where f isthe function you are zeroing, and x is your current approximation of the zero.

• Calculate an approximation to the derivative of the function at this point,from

f ′(x) =f(x+ ε)− f(x− ε)

2ε

• Calculate a new approximation to the zero, from

x = x− f(x)/f ′(x)

• Update the number of steps that you have taken through the loop.

(f) Return to main

Function M-file f

(a) This function has one input argument, x.

(b) This function has one output argument, y.

(c) The relation is

y = x2 − sin(x)/x;

23. Temperature along an iron bar

Consider an iron bar 100 cm long. One end of it is fixed at 0◦C while the other end isfixed at 100◦C. You must plot the temperature as a function of the distance betweenits end-points by dividing the bar into 1000 elements and using only the followinginformation:

• The temperature of the elements at either end is fixed. That is,TBar(1) = 0; TBar(1000) = 100;

• Except for the fixed endpoints, the temperature in any element is determined asthe average of its neighbors. That is,TBar(i) = 0.5*(TBar(i-1) + TBar(i+1))

• The solution is obtained by iteration within a while loop until the average temper-ature in the bar, sum(TBar)/1000, changes by less than 0.01◦C from the averagetemperature in the bar from the previous iteration.


Write two versions of MATLAB code that solve this problem. Both versions will employa while loop which computes the new average temperature in the bar.

One version will have a for loop (internal to the while loop) over the elements of thebar. The other version may not employ any for loops, employing instead MATLAB’sarray indexing method: the colon (:) syntax.

As the starting condition for the while loop, assume that the bar is 0◦C everywhereexcept at the warmer endpoint, TBar(1000).

24. Code Creation

The MATLAB script M-file below calls a MATLAB function M-file that returns theminimum, maximum, and average value of a 2D array. The average value is the sum ofthe matrix elements divided by the number of elements in the array. In the box below,write the function. You are not required to provide any comment lines in the function.

clear all;

Nrows = 37; Ncolumns = 93;

a = rand(Nrows,Ncolumns);

[minValue,maxValue,avgValue] = minMaxAvg(a,Nrows,Ncolumns)

25. Code Debugging and Refinement

The MATLAB function M-file below called quadRoot.m was discussed in earlier in thischapter.

%File: quadRoot.m

function [root1,root2] = quadRoot(A,B,C)

%QUADROOT returns the two real roots of A*x^2 + B*x + C = 0

%

% The two roots are root1 = (-B + sqrt(B^2 -4*A*C))/(2*A)

% root2 = (-B - sqrt(B^2 -4*A*C))/(2*A)

%

% An error is detected if 4*A*C > B^2 (imaginary roots)

% A warning is issued if 4*A*C = B^2 (identical roots)

if 4*A*C > B^2

error(’Roots would be imaginary. Get real!’)

elseif 4*A*C == B^2

warning(’Roots will be the same, degenerate’)

end

root1 = (-B + sqrt(B^2 -4*A*C))/(2*A); % This is line 15

root2 = (-B - sqrt(B^2 -4*A*C))/(2*A); % This is line 16

11.9. PROBLEMS 309

Unfortunately, the Professor wrote an incomplete program! When the function isemployed as follows:

>> [r1,r2] = quadRoot(0,1,1)

the following error is reported by MATLAB.

Warning: Divide by zero in quadRoot.m at line 15

Warning: Divide by zero in quadRoot.m at line 16

r1 = NaN

r2 = -Inf

>>

A problem occurs when A is zero. When A is zero there should be only one solution,x = -C/B unless B is zero as well. If both A and B are zero, there is no solution at all!Provide a fix-up to these problems by modifying the if/elseif/else construct above.

26. John Conway’s game of life

Write a MATLAB script M-file that plays John Conway’s game of life on a 50 by 50grid. Here is how to initialize the grid:

• Make a 50 by 50 matrix called grid of random numbers between 0 and 1.

• Consider each cell in grid. If grid(i,j) is less than 0.4, set it to one. Otherwise,set it to zero.

Definitions:

• A cell that contains “0” is “dead”.

• A cell that contains “1” is “alive”.

Here’s how the game is played:

(a) Make an exact copy of grid into the array called oldGrid.

(b) Loop through every cell in oldGrid.

(c) Find the number of neighbors a cell has by summing the values of the eightneighboring cells (left, right, up, down, and the four diagonal ones).

(d) If the cell is alive and it has less than 2 neighbors, it dies due to loneliness. (Setgrid(i,j) to zero.)

(e) If the cell is alive and it has 4 or more neighbors, it dies due to overcrowding.(Set grid(i,j) to zero.)

(f) If the cell is dead and it has exactly 3 neighbors, it comes to life. (Set grid(i,j)to one.)


(g) Any other condition, the cell stays as it was.

(h) After you have considered every cell, go back to step 1.

You do not have to write any code that produces graphical output. You just have tocode the algorithm described above.

Note that this program never terminates. Life goes on...

11.9. PROBLEMS 311

27. Slice this one up

Write MATLAB code that generates the 64 × 64 array that produces the followingplot.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

10 20 30 40 50 60

10

20

30

40

50

60

You may use the following commands to plot the array:

plotIt = b; % b is the array that contains the data to be plotted

plotIt(65,:) = 0; % Increase the number of rows by 1

plotIt(:,65) = 0; % Increase the number of columns by 1

pcolor(1:65,1:65,plotIt); % Plot it

colormap(gray)

colorbar

Try to make a solution that uses no more than 2 for-loops.

The array contains the data 0, 1, 2 in a pattern that produces the above figure. The64× 64 array can be made up of repeated use of an 8× 8 array.


28. Any way you slice it

Write two MATLAB codes that generate the 13×41 array that produces the followingplot. The first version uses for-loops only. The second version contains no for-loopsbut uses slice indexing of the form iStart:Stride:iEnd.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

5 10 15 20 25 30 35 40

2

4

6

8

10

12

14

You may use the following commands to plot the array:

plotIt = a; % a is the array to be plotted

plotIt(14,:) = 0; % Increase the number of rows by 1

plotIt(:,42) = 0; % Increase the number of columns by 1

colormap(gray)

pcolor(plotIt) % Plot it

shading flat

colorbar

Some helpful hints:

(a) The vertical bars are 1 unit wide and created with a Stride of 4.

(b) Note that the rectangle is in front of one of the vertical bars and in back of twoothers.

(c) The color of the vertical bars and edges are established by giving them a numericalvalue of 1.0.

11.9. PROBLEMS 313

(d) The color of the background is established by giving it a numerical value of 0.0.

(e) The color of the inner rectangle is established by giving it a numerical value of0.7.

29. Array striping

Write the MATLAB code that generates the array that produces the following plot. Itshould work for any response to user inputs (see example below).

0 5 10 15 20 25 30 35 400

5

10

15

20

The array contains the data 0 (black stripe) or 1 (white stripe) in a pattern thatproduces the above figure. The following response to the prompts produced the outputshown:

Number of stripes: 8

Number of Rows: 24

>>

30. Basket weaving

Write the MATLAB code that generates the array that produces the following basket-weave plot.

The array contains the data 0 (background), 0.6 (vertical dark grey stripes), or 1(horizontal light grey stripes) in a pattern that produces the above figure. The followingresponse to the prompts produced the output shown above:


How many times does the pattern repeat on each side? 5

>>

Hint: The solution is made a lot easier when you realize that there is a small 4 × 4array that is repeated to make the full pattern. Look at grid(1:4,1:4).

31. Minor diagonals and crosses

The built-in function, eye(M,N) creates anM×N matrix with 1’s on the main diagonaland 0’s elsewhere. Example:

>> eye(3,4)

ans =

1 0 0 0

0 1 0 0

0 0 1 0

(a) Write a function called littleEye() that does the same thing except the 1’s areon the minor diagonal. Example:

>> littleEye(3,4)

ans =

0 0 0 1

11.9. PROBLEMS 315

0 0 1 0

0 1 0 0

You should do this by having littleEye() call eye() and manipulating arrayindices using slice indexing.

(b) Write another function called cross() that sums eye() and littleEye(), exceptthat the maximum element returned can not be greater than 1. Example:

>> cross(3,3)

ans =

1 0 1

0 1 0

1 0 1

You can avoid using for loops by using the min() function. For example, if A isa scalar and B is a matrix, C = min(A,B) returns a matrix C that has no valuegreater than A.

32. A smooth operation

Consider the following MATLAB script M-file that calls the function called smooth.

NH = input(’Horizontal size of plot: ’);

NV = input(’Vertical size of plot: ’);

a = rand(NV,NH);

avgChng = 1;

while(avgChng > 0.001)

[a,avgChng] = smooth(a);

plotIt = a;

plotIt(NV+1,:) = 0; % Increase the number of rows by 1

plotIt(:,NH+1) = 1; % Increase the number of columns by 1

pcolor(0:NH,0:NV,plotIt)

colormap(gray)

colorbar

pause(0.01)

end

When a user types lookfor smooth the following is received:


>> lookfor smooth

SMOOTH smooth Smooth a 2D array

>>

When a user types help smooth the following is received:

>> help smooth

SMOOTH smooth Smooth a 2D array

The smooth function smoothes a 2D array, its only input parameter, by

performing adjacent horizontal and vertical point averaging following the

prescription:

Assuming:

R is the number of rows in the array

C is the number of columns in the array

When 2 <= i <= R and 2 <= j<= C

b(i,j) = (a(i-1,j) + a(i+1,j) + a(i,j-1) + a(i,j+1))/4

However, when the (i,j)’th point is in the first or last row or column of

the array, the algorithm averages only over the available horizontal or

vertical array points.

The smoothed array is returned to b, the first returned variable.

The average change is returned to the second return variable using the

following algorithm:

change = sum(sum(abs(b - a)))/(R*C)

>>

Write the function smooth.m. Your function has to respond to the lookfor and help

functions as described above.

33. Roots of a cubic equation

The roots (values of x such that f(x) = 0) of the cubic equation:

f(x) = Ax3 +Bx2 + Cx+D

11.9. PROBLEMS 317

where A, B, C and D are scalars, are given by the procedure below (attributed toFrancois Viete, 1615). You will code this algorithm into a MATLAB function calledcubeRoots.m where A, B, C and D are arrays:

• If A = 0 the above is not a cubic equation, it is a quadratic equation. Assumethat there exists a function called quadraticRoots.m somewhere on the MATLABpath that operates on arrays in the same way that cubicRoots.m does and willreturn real roots if they exist and NaN’s otherwise. Else,

• Define a, b, c as follows: a = B/A, b = C/A, c = D/A.

• Define Q = (a2 − 3b)/3, R = (2a3 − 9ab+ 27c)/54.

• If R2 > Q3 there are no real roots and so the 3 roots are NaN’s. Else,

• Define θ = cos−1(R/√Q3). (The MATLAB acos(u) function returns an array

the same dimensions as u but with its elements being the cos−1 of the elementsof u.)

• The three roots are:

x1 = −2√Q cos

(θ

3

)− a/3

x2 = −2√Q cos

(θ + 2π

3

)− a/3

x3 = −2√Q cos

(θ − 2π

3

)− a/3

where the MATLAB cos(u) and sqrt functions return an array the same di-mensions as u but with its elements being the cos and √ of the elements of

u.)

Some notes:

• Your function cubeRoots.m must accept four input arrays, A, B, C and D, allwith identical dimensions. Consequently, you must be careful with your use ofMATLAB operators.

• Your function returns three arrays containing either NaN’s or roots of the cubicequation.

• Your function must respond sensibly to the MATLAB lookfor and help func-tions.

• To make a variable into a NaN, assign it as follows: root = NaN;

34. Random, Rotated, Really

Write a MATLAB code that executes the following 5-step pseudocode. Each step isillustrated by a working example.


(a) Using the input() function, interrogate the user for R and C, a row size andcolumn size, respectively. Example:

Number of rows? 4

R =

4

Number of columns? 3

C =

3

(b) Using the input() function, generate a R × C random array. Example:

a =

0.0153 0.4660 0.2026

0.7468 0.4186 0.6721

0.4451 0.8462 0.8381

0.9318 0.5252 0.0196

(c) Rotate the array counter-clockwise. The first column becomes the last row, thesecond column becomes the second-to-last row and so on. Example:

a =

0.2026 0.6721 0.8381 0.0196

0.4660 0.4186 0.8462 0.5252

0.0153 0.7468 0.4451 0.9318

(d) Set to zero any array element that has a greater column index than row index.Example:

a =

0.2026 0 0 0

0.4660 0.4186 0 0

0.0153 0.7468 0.4451 0

(e) Change the sign of any array element that has a value less than 1/2. Example:

a =

-0.2026 0 0 0

-0.4660 -0.4186 0 0

-0.0153 0.7468 -0.4451 0

35. Can you (co)relate?

Write a MATLAB function M-file called correlate.m that does the following:

• Accepts a 2-dimensional array of doubles of any size. This array is the onlyvariable in the input list.

11.9. PROBLEMS 319

• Returns the smallest absolute difference that exists between any two array ele-ments.

• Returns the indices of the two closest elements.

• Responds sensibly to the lookfor and help functions

• Note: abs(x) returns the absolute value of all the elements of x.

Here’s a working example:

>> lookfor correlate

CORRELATE correlate, finds the smallest difference between two array elements

>> help correlate

CORRELATE correlate, finds the smallest difference between two array elements

Returns the difference and array indices of the two closest elements

Programmer: Alex Bielajew 12/12/00

Problem on Final Exam for Eng101/200, Fall ’00

>> A = rand(3,4)

A =

0.9501 0.4860 0.4565 0.4447

0.2311 0.8913 0.0185 0.6154

0.6068 0.7621 0.8214 0.7919

>> [d,i1,j1,i2,j2] = correlate(A)

d =

0.0086

i1 =

2

j1 =

4

i2 =

3

j2 =

1

36. Minesweeper initialization

Consider the MATLAB script M-file started below:


%MINESWEEPER minesweeper: Initialize the minesweeper game

clear all

echo off

clc

R = input(’Number of rows in the grid: ’);

C = input(’Number of columns in the grid: ’);

nMines = input(’Number of mines: ’);

if (nMines < 1)

error(’Too few mines’)

elseif (nMines > R*C)

error(’Too many mines’)

end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

% Your part goes here.

% Write your code on the following page(s).

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

disp(grid)

Here’s is what you must do.

(a) Put nMines randomly on the grid by giving that grid point the value -1. Forexample, grid(i,j) = -1 means that the i,j’th point on the grid contains a”mine”. You will need to use both the rand and int32 functions. (int32()converts its argument to a 32-bit integer that may be used as an array index.) Besure that you deposit exactly nMines mines on the grid, no more, no fewer.

(b) For each grid(i,j) point that is not a mine, count the number of mines sur-rounding it, those that are in the range grid(i-1:i+1,j-1:j+1). Be carefulwith the boundaries of the grid. Note: sum(sum(a)) sums up all the values onthe points of the array called a.

(c) Here is an example use of the program:

>> minesweeper

Number of rows in the grid: 5

Number of columns in the grid: 3

Number of mines: 5

1 -1 1

11.9. PROBLEMS 321

1 2 2

0 2 -1

1 4 -1

-1 3 -1

>>

37. This problem is a minefield

Consider the MATLAB script M-file started below:

%LANDMINE landmine: A game of landmine

clear all

echo off

clc

R = input(’Number of rows in the grid: ’);

C = input(’Number of columns in the grid: ’);

pMine = input(’Probability that a tile contains a mine: ’);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

% Your part goes here.

% Write your code on the following page(s).

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

showGrid = grid;

showGrid(R+1,:) = -1;

showGrid(:,C+1) = 1;

pcolor(0:C,0:R,showGrid)

header = sprintf(’\\fontsize{20}There were %d survivors\n’,nSurvived);

title(header)

axis off

This is a brief description of what you must do.

(a) Between 1:R and 2:C-1 use the random number function rand to decide whetheror not that gridpoint contains a landmine. In other words, if rand < pMine setthe grid point to -1 to indicate the presence of a mine. Otherwise, set it to 0, toindicate that the gridpoint is safe.


(b) Make the first column of the grid equal to 1. A “1” indicates a person on thegrid. These people are trying to make it to the last column on the grid withouttouching a mine.

(c) Within a while loop whose logical condition is that there are still people withingrid(:,1:C-1), use a double for loop to look at every point within grid(:,1:C-1).If that point has value 1, then do the following:

i. Attempt to advance the person one square to the right (horizontally) andwith equal chance: one square up, one square down, or no vertical change.Restrict the vertical movements so that the person stays within the grid. Useof the max() and min() functions, as described in class and in an earlier examproblem, makes this restriction quite efficient.

ii. If the proposed new location contains another person, then nothing happens.

iii. If the proposed new location contains a landmine, then both the person andthe landmine are eliminated.

iv. If the proposed new location is within the last column, then that person hassafely reached home and should not move anymore.

(d) The game continues until there are no people remaining within grid(:,1:C-1)

(e) Be careful not to move a person more than once during each execution of thedouble for loop.

38. War is heck

Write a MATLAB code for the game of War that is described as follows:

• The game is played on an R × C array, where R and C are inputs provided by theuser.

• The array elements are filled with random numbers using the rand() function.

• Go through each element of the array. If the element is greater than 1/2 reset thearray element to +1, else, set it to -1. Each one of these elements is a player in thegame. Those assigned +1 are the Positives. Those assigned -1 are the Negatives.Both the Positives and Negatives think of each other not only as Opposites butalso as Opposition—deadly opposition.

• Print out the starting number of Positives and Negatives.

• The Positives and Negatives are at war and here is how the battle is fought.

• An outer while loop is executed under the condition that there are players of bothkinds still in the game.

• During each execution of the while loop, the following happens:

(a) Each player tries to moves randomly but with equal chance in one of the 4directions, North(N), South(S), East(E) or West(W) by one array index.

11.9. PROBLEMS 323

(b) If a player wants to move outside the bounds of the array, then the playerdisappears from the game. The penalty for cowardice is severe. (Set thatarray element to 0.)

(c) If the array element that the player wants to move to is empty (0), the playermoves there. (The +1 or -1 gets written into the new array element. The oldarray element is set to 0.)

(d) If the array element that the player wants to move to is a player of the sametype, then nothing happens.

(e) If the array element that the player wants to move contains an Opposite, thenthey fight. Three things can happen, all with equal chance:

i. Mutual destruction—both players perish. (Both array elements get setto 0.)

ii. The Positive player wins and the Negative player is eliminated from thegame. (The new location gets set to +1 and the old location gets set to0.)

iii. The Negative player wins and the Positive player is eliminated from thegame. (The new location gets set to -1 and the old location gets set to0.)

• When the while loop ends, there are 3 possible outcomes. Provide a message thatindicates how the game ended and the number of remaining players.

Here’s an example of running the code.

>> war

Number of rows in the game: 10

R =

10

Number of columns n the game: 12

C =

12

The game starts with 64 +’s and 56 -’s

The -’s have annihilated the +’s. There are 17 -’s remaining.

>>

39. Gaussian sum Write a MATLAB function M-file, called gauss(), that is called asfollows:

s = gauss(N) % where N in some integer

Your gauss.m responds to the lookfor and help functions as follows:


>> lookfor gauss

GAUSS gauss The Gaussian sum: gauss(N) = 1 + 2 + 3 + ... + N

>> help gauss

GAUSS gauss The Gaussian sum: gauss(N) = 1 + 2 + 3 + ... + N

>>

which gives the algorithm. Feel free to use the “trick” if you remember it.

40. Fibonacci sum

Write a MATLAB function M-file, called fibonacci, that is called as follows:

s = fibonacci(N) % where N is some integer

Your fibonacci.m responds to the lookfor and help functions as follows:

>> lookfor fibonacci

FIBONACCI fibonacci Returns the fibonacci number F(N)

>> help fibonacci

FIBONACCI fibonacci Returns the fibonacci number F(N)

The Fibonacci numbers are defined recursively thus:

F(1) = 1; F(2) = 1; F(N) = F(N - 1) + F(N - 2) where N >= 3

which gives the algorithm. The error function is used to return a message when theparameter (N in the call statement above) is less than or equal to 0. For example...

>> fibonacci(0)

??? Error using ==> fibonacci

Fibonacci numbers are defined for N >= 1

You may not use recursive function calls.

41. Array striping

Complete the MATLAB code below that generates the array that produces the follow-ing plot, in response to the following request for inputs:

Number of horizontal (Gaussian) white stripes (>0): 10

Number of vertical (Fibonacci) white stripes (>2): 8

Your code should work for any positive (> 0) response to the prompts.

The array contains the data 0 (black rectangles) or 1 (white stripes) in a patternthat produces the above figure. The i’th vertical stripe is situated at the indexfibonacci(i) (See preceding problem.) while the i’th vertical stripe is situated atthe index gauss(i) (See two problems back.) You may assume (if you wish) that thegauss() and fibonacci() functions, as described in the 2 previous problems, are onthe Matlab search path.

11.9. PROBLEMS 325

0 2 4 6 8 10 12 14 16 18 200

5

10

15

20

25

30

35

40

45

50

55Black has value 0, White has value 1

NH = input(’Number of horizontal (Gaussian) white stripes (>0): ’);

NV = input(’Number of vertical (Fibonacci) white stripes (>2): ’);

%Add your MATLAB code starting here...

%The code below plots the graph if you supply

%NRows, NCols and a(1:NRows,1:NCols)

plotIt = a;

plotIt(NRows+1,:) = 0; % Increase the number of rows by 1

plotIt(:,NCols+1) = 0; % Increase the number of columns by 1

pcolor(0:NCols,0:NRows,plotIt)

title(’\fontsize{20}Black has value 0, White has value 1’)

colormap(gray)

42. Waldo wanders again

Write a simplified version of the Waldo program you encountered in Chapter 5. Hereare the rules:

• Waldo’s universe is a = zeros(20)

• Waldo starts off at a(10,10) = 1


• As indicated above, the indices of the array where Waldo is has a value 1, and 0otherwise.

• Waldo is trying to make it to a(16:20,16:20). If he makes it there, the game isover and Waldo has found his way home.

• On each step, Waldo attempts to walk randomly into one of the 8 adjacent squares,up, down, left, right and the 4 diagonal ones.

• If Waldo attempts to walk out of a(1:20,1:20), Waldo is lost and the game isover.

• You do not need to supply any graphical output.

11.10. PROJECTS 327

11.10 Projects

1. Is your dorm room cold?—Register your complaint!

Consider that your dorm room is represented by a 32×32 array, with 4 walls fixedat 20◦C, a window 16×1 centered on the left wall which is fixed at 0◦C (it’s freezingoutside), and a heat register comprised of 8 contiguous pixels (array elements that alltouch another heat register array element on at least one side) that is fixed at somehigher temperature. See Figure 1 for an example. Your job is to:

(a) Locate the register in the room somewhere (not occupying the same space as awall or a window!) so that the temperature in the room is as uniform as possible,

(b) Heat the room to between 21 and 22◦C by raising the temperature of the register.

(c) Full marks for algorithm design will be awarded to those who find a location forthe heat register that keeps the temperature in the room uniform to within 2degrees. (How to calculate this is discussed later.)

Here is a step-by-step procedure for this problem:

(a) Open a file called a7.m with any editor. This will be your script M-file. Oncein MATLAB, you will run this program by simply typing a7 at the MATLABprompt.

(b) To start the problem, first fill a 32×32 array with zeros:

N = 32;

T = zeros(N);

(c) Then, make the first and last pixels of each row and column equal to 20, settingthe temperature on the walls, then the 16 pixels of the window (fixed at 0) andfinally the 8 pixels of the register (set at T = 65◦C) as shown in Figure 1.

The picture in Figure 1 is displayed using the following MATLAB commands:

plotT = T; % Copy the T array

plotT(N+1,:) = 0; % Increase the number of rows by 1

plotT(:,N+1) = 0; % Increase the number of columns by 1

pcolor(1:N+1,1:N+1,plotT); % Plot it

colorbar;

Increasing the row and columns by 1 is necessary to get MATLAB to show thewalls and the window, otherwise only the interior of the room would show.

(d) Inside a while loop, the i,j’th cell gets it value from its neighbors, via an averagingrelation of the form:

T(i,j) = (oldT(i+1,j) + oldT(i-1,j) + oldT(i,j+1) + oldT(i,j-1))/4;


where oldT is the temperature array after the previous pass through the while

loop. You have to program this in the appropriate place in the while loop and youhave to make sure that you do not violate the boundaries of the array. You canuse for loops to do this although it is much more efficient to use “slice indexing”.

(e) In the above averaging procedure, you may have overwritten some of the arraythat was meant to be fixed by the initial conditions, like the register, for instance.So, you had better make sure that you fix these up in the appropriate place beforethe while loop executes another pass.

(f) The while loop is executed until the average change temperature changes lessthan 0.00001. The average change is calculated from:

difference = abs(T - oldT);

change = sum(sum(difference))/(N^2);

(Do a “help” on abs and sum to find out what these functions do.)

(g) If you get everything correct, the final results look like Figure 2 which is plottedusing the commands pcolor(1:N,1:N,T); shading interp; colorbar;.

(h) Calculate and print out the average temperature and its variation:

average = sum(sum(T))/(N^2);

variation = sum(sum(abs(T - average)))/(N^2);

For the above example, I got the average to be T avgfinal = 21.1514 while the variationwas ∆T avgfinal = 5.39824. You should verify that you get this result before continuing.While this got the average temperature to within specifications, the variation isway out of spec. You have to do better. The easiest but most boring way to do it isto move the register around by typing in the configuration in the script M-file andrunning it again until you satisfy both the average and variation specifications.(BLIF’s rule #197 of computer programming: Always do it the easy, boring wayfirst!) In this case, however, you will need some intuition to get the answers towithin specifications because there are many possibilities, both in the location ofthe register and its shape. You should have a look in your dorm or other roomsto see if you find any common relationships between placement of windows andheat registers. You should also hope that the architects and engineers thoughtabout this carefully because the impact on heating costs and human comfort arequite large.

Alternatively, you can try to find a way of automating the optimization procedure.How you do this is up to you! Be creative!

Finally, Figure 3 shows a different starting position with the ending results shown inFigure 4. This is a terrible solution! Not only is the variation worse, but the registeris so hot that you would burn yourself if you got close to it!

11.10. PROJECTS 329

0

10

20

30

40

50

60

5 10 15 20 25 30

5

10

15

20

25

30

Figure 11.1: Typical starting condition for the 32×32 room with a register on the right wallat T = 65◦C.


0

10

20

30

40

50

60

5 10 15 20 25 30

5

10

15

20

25

30

Figure 11.2: Final result for a 32×32 size room starting with the configuration of Figure 1.In this case, T avgfinal = 21.1514 and ∆T avgfinal = 5.39824.

11.10. PROJECTS 331

0

10

20

30

40

50

60

70

80

90

100

5 10 15 20 25 30

5

10

15

20

25

30

Figure 11.3: Another starting condition for the 32×32 room with a register in one of thecorners at T = 105◦C.


0

10

20

30

40

50

60

70

80

90

100

5 10 15 20 25 30

5

10

15

20

25

30

Figure 11.4: Final result for a 32×32 size room starting with the configuration of Figure 3.In this case, T avgfinal = 21.1643 and ∆T avgfinal = 6.1069.

11.10. PROJECTS 333

2. The plasma arc lamp—Do you get the point?

If you take two electrically conducting objects and connect then to opposite poles of abattery, nothing happens—usually. However, if you touch the objects together you canget a spark. If the voltage difference is great enough, you do not need to touch themtogether to have something interesting happen. Two conductors, if connected across avoltage that is high enough and brought close enough together, will cause an “arc” tobe produced. This is lightning on a small scale!

If causing an “arc” is desirable for some reason, it helps to have the conductors“pointed”. This is why lightning rods are pointed—to attract the lightning in anefficient manner and then divert it into the ground safely away from things that canbe damaged by it.

You can also make a brilliant light using high voltage and pointed conductors. Thesedevices are called “plasma arc lamps” or “plasma arc torches”. The point (pun in-tended) of this problem is to make a plasma arc lamp, at least a virtual one, on thecomputer. The pictures that are produced are a thing of beauty, well, at least to me.

The template for this problem is found at the end of this problem description.

Here is a step-by-step procedure for solving this problem:

(a) Type the template into a file.

(b) Start MATLAB and run the template program. The first thing that pops up is agraphical representation of the start to the problem. It should look like:

−1000

−800

−600

−400

−200

0

200

400

600

800

1000

5 10 15 20 25 30

5

10

15

20

25

30

Figure 11.5: Starting conditions for a 32×32 voltage grid.


Do not proceed unless you get this.

The initial conditions for the voltage in a square grid of points (the grid size mustbe a multiple of 32) is determined as described below. The statement:

V = zeros(N);

creates an N x N array to represent the voltage and sets its values to 0. Thestatement:

V(1:15*N/32,1:15*N/32) = 1000;

creates the high voltage (+1000 Volts) electrode in the lower left hand corner ofthe figure, while the statement:

V(17*N/32+1:N,17*N/32+1:N) = -1000;

creates the low voltage (-1000 Volts) electrode in the upper right hand corner ofthe figure. The statements:

V(1,15*N/32+1:N) = linspace(1000,0,N - 15*N/32);

V(15*N/32+1:N,1) = linspace(1000,0,N - 15*N/32)’;

V(N,1:17*N/32) = linspace(0,-1000,N - 15*N/32);

V(1:17*N/32,N) = linspace(0,-1000,N - 15*N/32)’;

establish a linear change in the voltage from the electrodes to the upper left andlower right hand corners, so that these points at the upper left and lower right arezero. Note the use of the transpose operator “’” in two of the above statements.

(c) If you then type a <RETURN> in the MATLAB command window, the figure onthe previous page is replaced by a shading interp version and a second windowpops up which will be empty. Now you have to start some programming.

(d) Inside the while loop, the i,j’th cell gets it value from its neighbors, via a relationof the form:

V(i,j) = (oldV(i+1,j) + oldV(i-1,j) + oldV(i,j+1) + oldV(i,j-1))/4;

where oldV is the voltage array after the previous pass through the while loop.You have to program this in the appropriate place in the while loop and you haveto make sure that you do not violate the boundaries of the array. You can usefor loops to do this although it is much more efficient to use “slice indexing” asin the determination of the initial conditions above.

(e) In the above averaging procedure, you may have overwritten some of the arraythat was meant to be fixed by the initial conditions. So, you had better makesure that you fix these up in the appropriate place in the while loop.

(f) The while loop is executed until the voltage array does not change by a significantamount. Code to test for this is provided.

11.10. PROJECTS 335

(g) After the voltage array is calculated, you calculate the magnitude of the electricfield from it. The magnitude of the x-component of the electric field is obtainedfrom:

Ex(i,j) = V(i+1,j) - V(i,j);

while the magnitude of the y-component of the electric field is obtained from:

Ey(i,j) = V(i,j+1) - V(i,j);

The magnitude of the electric field is obtained from a relation of the form

E(i,j) = sqrt(Ex(i,j)*Ex(i,j) + Ey(i,j)*Ey(i,j));

Be very careful of the array bounds!

(h) Then the magnitude of the electric field is plotted as provided by the template.

(i) If you get everything correct, the final results look like:

20 40 60 80 100 120

20

40

60

80

100

120

Figure 11.6: Final result for a 128×128 voltage grid.


20 40 60 80 100 120

20

40

60

80

100

120

Figure 11.7: Electric field grid corresponding to the voltage in the previous figure.

11.10. PROJECTS 337

58 60 62 64 66 68 70 7256

58

60

62

64

66

68

70

Figure 11.8: A “zoom in” on the previous picture to show the details of the arc. Even on afine 128×128 grid you can see evidence of the “digitization” of the grid.


The code template...

N = 32; % Array size, should be divisible by 32: 32, 64, 96, 128,...

V = zeros(N); % Creates Voltage array and sets its values to 0

% Sets initial values of the V array

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

V(1:15*N/32,1:15*N/32) = 1000; % High voltage "point"

V(17*N/32+1:N,17*N/32+1:N) = -1000; % Low voltage "point"

% Linear ramps along edges from the high voltage point

V(1,15*N/32+1:N) = linspace(1000,0,N - 15*N/32);

V(15*N/32+1:N,1) = linspace(1000,0,N - 15*N/32)’;

% Linear ramps along edges from the low voltage point

V(N,1:17*N/32) = linspace(0,-1000,N - 15*N/32);

V(1:17*N/32,N) = linspace(0,-1000,N - 15*N/32)’;

% Plot the initialized array

%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% The following lets all the cells of V be printed,

% useful for making sure the array got initialized OK.

% Without this, the N’th row and column of V would not

% be displayed.

% Should NOT do it this way if using shading interp

plotV = V; % Copy the V array

plotV(N+1,:) = 0; % Increase the number of rows by 1

plotV(:,N+1) = 0; % Increase the number of columns by 1

pcolor(1:N+1,1:N+1,plotV); % Plot it

colorbar;

pause; % Wait for a keyboard event before continuing

tic; % Starts the clock

oldV = V; % Save the initial value of the array in "oldV"

% Enter a loop that ends when the change in voltage is

% equal to or less so some value

11.10. PROJECTS 339

change = 1; % Sets change to 1 to get the while loop started

while(0.001 <= change)

% Take the average voltage of the 4 adjacent cells

%...need student input here...

% Resets initial values that should stay constant (see above)


% Calculates the change in voltage

difference = abs(V - oldV);


% Prints out the change in voltage

fprintf(’Average change = %g\n’,change)

% Gives the oldV array the newly calculated voltage

oldV = V;

end

time = toc; % Stores how long the program took to run

fprintf(’Elapsed time = %g\n’,time) % Prints it out

% Plots a representation of the voltage

pcolor(1:N,1:N,V);

shading interp;

colorbar;

% Calculate the magnitude of the electric field

Ex = zeros(N-1,N); % Allocate memory for Ex

Ey = zeros(N,N-1); % Allocate memory for Ey

E = zeros(N-1,N-1); % Allocate memory for E


figure(2);

pcolor(1:N-1,1:N-1,E);

shading interp;


3. Battleship revisited

You will write MATLAB code that will allow you to play the game “Battleship!” onyour computer. There is only one ship, the size of the ship is 4 units and you onlyhave 25 bombs. Otherwise the game rules are exactly the same as for the Battleshipprogram discussed in Chapter 8.

However, you must use graphical output to show the status of the game board (withthe ship hidden) and use the mouse to generate the bomb drops.

Hints:

Initializing the game grid

N = 10;

showBoard = zeros(N); % Make the board, value of zero for "open sea"

For more explanation:

>> help zeros

Drawing the game grid If you have an array called showBoard, here is how to dis-play it:

plotBoard = showBoard; % Makes a copy of showBoard into plotBoardplotBoard(N + 1,:) = 0; % Boosts number of rows in plotBoard by 1, assigns 0’splotBoard(:, N + 1) = 0; % Boosts number of columns in plotBoard by 1figure(1); % Working with figure 1pcolor(0:N,0:N,plotBoard); % Make the plotaxis off % Turn of axis labeling (may want to leave on during debugging)


>> help figure

>> help pcolor

>> help axis

Choosing the first point of the ship

i = 1 + N*rand;

j = 1 + N*rand;

hiddenBoard(int32(i),int32(j)) = 1; %Ship point has value 1

Note that you will have to have two boards, one that you show during the gameand one that is kept hidden, where the ship points reside.

Note the use of the int32() function. The indices of an array should always beintegral. The statements i = 1 + N*rand; j = 1 + N*rand; are intrinsicallydoubles. You need the int32() function to convert from doubles to integers.


11.10. PROJECTS 341

>> help rand

>> help int32

Dropping bombs with the mouse

[x,y] = ginput(1);

jDrop = int32(x + 1);

iDrop = int32(y + 1);

The above generates i,j indices on the game board. You can testhiddenBoard(iDrop,jDrop) to see if that array elements represents open sea(value 0), a part of the ship (value 1), or something else. In my version of thegame, I used a value of 0.25 to represent a bomb dropped on open sea, and a 0.75to represent a struck ship.


>> help ginput


4. Swarm ball

This project was inspired by my 7-year-old daughter’s soccer team, The Muggles. If youhave ever watched a group of 7-year-old’s play soccer, it bears very little resemblance tothe game called soccer. It resembles more closely what would should be called “swarmball”.

In swarm ball, the object of the game is to surround the ball with as many of your5 other teammates as possible, for the chances of moving the ball in the direction ofthe opponents’ goal is proportional to the number of your team surrounding the ball.Of course, the other team has a similar strategy. So, around the ball is a swarm ofplayers all kicking at the poor thing. The ball usually blurts out in one direction andthe swarm reassembles around the ball. Goaltenders, while still a part of swarm ball,are almost completely irrelevant. The ball is always rolling on the ground. Would youbend down to pick up a soccer ball surrounded by 24 flailing feet? I think not.

So, this project’s task is to program a simulation of swarm ball with the followingrequirements.

(a) A 2-dimensional array, called field, contains the game, the players, the ball andthe goalposts. The width and length of the field are defined by:

fieldWidth = 31; % Width of the field

fieldLength = 41; % Length of the field

(b) The following values of the elements of field determine what that point occupies.Use the following as it gives a sensible rendition when field is plotted usingpcolor:

grass = 0.5; % Grass color

ball = 0.62; % Soccer ball’s color

red = 0.8; % Red’s team color

blue = 0.2; % Blue’s team color

redGoal = 0.7; % Red’s goal color

blueGoal = 0.4; % Blue’s goal color

boundary = 0; % Field boundary

(c) Define the field, the boundaries and the goals as follows:

field = zeros(fieldWidth,fieldLength);

field(2:fieldWidth-1,2:fieldLength-1) = grass;

field(10:22,1) = redGoal;

field(10:22,end) = blueGoal;

(d) Define the starting position of the game using the following indices:

ballX = 21;

ballY = 16;

11.10. PROJECTS 343

field(ballY,ballX) = ball; % The ball’s starting position

% Red team starting positions

redX = [17,17,17,11,11,11];

redY = [13,16,19,10,16,22];

% Blue team starting positions

blueX = [25,25,25,31,31,31];

blueY = [13,16,19,10,16,22];

and then place these points on the field.

(e) The game proceeds in a series of steps, not to exceed 120, when time runs out.During each step, the following happens:

i. Advance each player towards the ball in either the vertical, horizontal ordiagonal directions. A player can only move to an open point (no otherplayers, boundaries, goals or ball). Be sure that you do this in a fair fashion.If you do not and move, say, all the red players first, they would block theblue players from the ball and the red team would have an unfair advantage.

ii. Count the number of players (redKick, blueKick) of each team surroundingthe ball. You must check all 8 points, vertical, diagonal and horizontal, inthe vicinity of the ball. The team that kicks the ball will be determined bythe following prescription:

% Pick which team gets the kick

if (redKick == 0 & blueKick == 0)

kick = 0; % Neither team kicks

elseif (rand < redKick/(redKick + blueKick))

kick = 1; % Red team kicks

else

kick = -1; % Blue team kicks

end

iii. Pick the direction that the ball will go according to the following prescription:

% Pick a direction to kick the ball

if (kick)

% Add some randomness in the y direction, move towards the goal

randKick = rand;

if (randKick <= 0.3333)

kicky = -1;

elseif (randKick <= 0.6667)

kicky = 0;

else

kicky = 1;

end


% Make sure kick is usually in the correct direction

kickx = 0;

if (rand < 0.75)

if (kick > 0)

kickx = 1;

elseif (kick < 0)

kickx = -1;

end

end

if (kickx == 0 & kicky == 0)

kick = false; % No kick, a total miss

end

end

iv. When the kick direction is determined, place the ball in the first availablepoint along the direction. If ends up in a goal, one of the teams wins and thegame is over. If the ball hits the boundary, there is no kick. (This is simplerto program than a throw-in from the sidelines, or a goal kick.)

(f) Use a pcolor display after each repositioning and each kick.

11.10. PROJECTS 345

5. Going, going, going...Gone!

Baseball, like most sports, is a game of inches. (Except for soccer, of course, which isa game of feet!) However, some things in baseball happen on a grand scale, like thedistance a ball travels when hit for a home run.

There are two parts to this problem, both to be solved with MATLAB. You can solvethem using any numerical/computational technique you like so long as it is done withMATLAB. You must use a function called homeRunDistance which is described in alittle detail below.

(a) A baseball is struck at a height of 3 feet about the ground. It can have any velocitybetween 0 (Bielajew territory) and 250 feet/second (McGuire/Sosa territory). Itcan have any angle with respect to the ground from 0 (parallel to the ground) to90 degrees (straight up in the air).

Using the equations of motion described below, write a MATLAB function thatstarts something like this. (It can be different.)

function d = homeRunDistance(y0,theta,v,G,D)

%HOMERUNDISTANCE returns distance a ball travels (hits the ground)

% homeRunDistance(y0,theta,v,G,D) calculates distance a baseball travels

% before it hits the ground assuming that:

% y0 is the initial height of the ball (ft)

% theta is the angle in degrees of the start of the trajectory with

% respect to the ground (90 is straight up)

% v is the initial speed (ft/s)

% G is the magnitude of the acceleration due to gravity (ft/s/s)

% D is the magnitude of the drag factor (1/ft)

%

% The components of acceleration are:

%

% ax = -D*vx*v (x-direction in units ft/s/s)

% ay = -D*vy*v - G (y-direction in units ft/s/s)

%

% where

%

% vx = v*cos(pi*theta/180) (x-direction velocity in units ft/s)

% vy = v*sin(pi*theta/180) (y-direction velocity in units ft/s)

Using a script M-file to call the function homeRunDistance, make a 3D plot ofdistance versus theta and v between the limits described above. (A 20 by 20array should be sufficient.)

(b) For an initial velocity of 250 ft/s, what angle should the ball be hit for maximumdistance? You can determine the maximum in any way you please as long as it is


part of the same script M-file used above and it gives you approximately the rightanswer! When you obtain this angle make a 2D plot of the trajectory of the ballboth with and without air resistance (D = 0). “Plotting the trajectory” meansmaking a 2D plot where the y-axis is the height and the x axis is the distance.Without air resistance it would have a parabolic shape.

11.10.1 Supplementary material: Trajectories without air re-sistance

If a ball starts off at height y0, horizontal position x0 = 0, with components of velocityvx,0 and vy,0 in the presence of constant gravity providing an acceleration towards theground, ay = −G, the situation is one of constant acceleration.

The trajectory as a function of time may then be described by:

x = vx,0t

y = y0 + vy,0t−1

2Gt2

We can solve for the time of flight, tflight, it takes for a ball to return to the groundusing the quadratic equation and obtain the time and distance as follows:

tflight =vy,0 +

√v2y,0 + 2Gy0

Gd = vx,0tflight

For example, if we use y0 = 3 (ft), vx,0 = vy,0 = 250/√2 (ft/s), and G = 32.174

(ft/s/s) we get d = about 1950 feet for the distance! The inputs are realistic and sowe know that air resistance must play a big role. However, knowing that there is anexact answer in the case of no wind resistance will serve as an important test of ourstepping algorithm described later.

11.10.2 Supplementary material: Trajectories with air resis-tance

In this case, the equations of motion are different because the accelerations are notconstant. In this case they are:

ax = −Dvx

√v2x + v2y

ay = −Dvy

√v2x + v2y −G

11.10. PROJECTS 347

whereD is a measure of the resistance of the air. The total air resistance is proportionalto v2 and points in the direction opposite to the instantaneous direction of the ball.

The drag factor is related to the physical characteristics of the ball (or projectile) andhas the form:

D =1

2CD,ballρairAball/mball

where ρair air is the density of the air, Aball is the cross sectional area of the ball, mball

is the mass of the ball and CD,ball is the “drag coefficient”, a unitless quantity. Forsmooth spheres, CD,ball is always very close to 0.5. However, for a baseball it’s closerto 0.35 because of the seams. A pitcher can also change CD,ball by adding a little extrato it (saliva, vaseline) or taking a little away (cutting the ball). Baseballs that arenewly manufactured have a shine that is rubbed off before the start of the game by theumpires and occasionally by the pitchers. This IS legal. All they are doing is makingthe CD,ball a little bit smaller to make the pitches faster.

The D for a major-league baseball can be worked out easily form the above relations.Use D = 0.0019285 and G = 32.174 for this problem.

11.10.3 Supplementary material: Stepping algorithms

Let’s go back to the equations of motion.

ax = −Dvx

√v2x + v2y

ay = −Dvy

√v2x + v2y −G

If we assume that the acceleration is constant over a small time interval, ∆t, then makea stepping algorithm as follows:

(a) Start with the initial conditions x0 = 0, y0 = 3, vx,0 = v0 cos θ and vy,0 = v0 sin θ,t0 = 0, and a step counter i initialized to 0.

(b) Increment the step counter i = i+ 1.

(c) Calculate the acceleration at the start of the step:

ax,i = −Dvx,i−1√v2x,i−1 + v2y,i−1

ay,i = −Dvy,i−1√v2x,i−1 + v2y,i−1 −G


(d) Update the positions and velocities assuming constant acceleration.

xi = xi−1 + vx,i−1∆t+1

2ax,i(∆t)2

yi = yi−1 + vy,i−1∆t+1

2ay,i(∆t)2

vx,i = vx,i−1 + ax,i∆t

vy,i = vx,i−1 + ay,i∆t

(e) Stop if yi falls below 0, otherwise go back to step 2.

There are a few subtleties you will have to work out. One is to guarantee that they never becomes negative. The other is to make a reasonable choice for ∆t. Justifyyour choice. Recall that we are assuming that the acceleration is constant over thestep. It isn’t. However, a proper choice for ∆t make sure that the acceleration isnearly constant. This week’s lectures will be riddled with hints on the solution of thisproblem!

11.10. PROJECTS 349

6. This problem sucks!

Have you ever wondered how one can make a vacuum?

One can make a good vacuum, comparable to the molecular/atomic density in deepspace by employing a multi-stage process. In the first stage would we would take, say,a cylinder with a piston in it (much like a combustion chamber in a car) and a pipeattached to a closed canister that you want to evacuate. With a valve in the canister(to the cylinder) open and a valve in the cylinder (to the outside) closed, withdrawthe piston. The pressure in the canister and the pipe is reduced. Now, close the valveto the canister and open the valve to the outside, allow the piston to descend andrepeat. One can only go so far with this method. Soon there are too few molecules inthe canister and enough molecules in the mechanism of the outside valve that one getsvery little or no gain on each stage of the pump cycle.

The second stage (the one that this problem is concerned about) is to introduce acold trap. A separate chamber (attached to the canister) has a cold surface in it towhich molecules adhere. This further reduces the number of molecules. A third stagemight involve causing the molecules to become ionized and then they may be “sweptout” with an electric field. Making a vacuum is an interesting combination of a lotof different technologies—mechanical, electrical, chemical... Let’s focus on the secondstage. With MATLAB, first construct two chambers of identical dimensions, connected

−1

−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

5 10 15 20 25 30

5

10

15

20

25

30

Figure 11.9: Two chambers and a pipe (evacuated, value = 0) surrounded by walls (solid,value = -1).

together with a pipe. The ones I made were made up of 2D arrays and look like thosedepicted in first figure. In my example I used chambers that were 30 units high and 10units wide. The pipe was 16 units high and 10 units wide. (A “unit” is the width of


a “cell”.) Thus the entire geometry of the pipe, the two chambers and enclosing wallscould be represented by a 32 × 32 array in MATLAB. The space within the walls isgiven the value “0” for vacuum and the walls are given the value “-1”.

Now, fill one of the chambers with molecules, placed randomly. In the example insecond figure, I placed 100 molecules. Note that I allowed more than 1 molecule percell. Each cell contains a number that represents the number of molecules within thecell. Since MATLAB’s random number generator seeds differently every time, yourdistributions will look different.

−1

−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

5 10 15 20 25 30

5

10

15

20

25

30

Figure 11.10: The left chamber filled with 100 molecules.

Now, a time-step is executed. During this time-step, each individual molecule can moverandomly in any direction by one cell, upwards, downwards, sideways, or diagonally.If the new direction would cause the molecule to hit one of the walls, the proposedstep is invalid and a new direction should be picked randomly. If a molecule entersthe rightmost column of cells, it “sticks” to the cold wall and can no longer be moved.The third figure shows the distribution of molecules after 2000 time-steps.

Finally, the simulation ends when all the molecules are trapped in the cold trap. Thefinal configuration of my example is shown in the fourth figure.

Here is what you must do to complete this problem.

11.10. PROJECTS 351

−1

0

1

2

3

4

5

6

5 10 15 20 25 30

5

10

15

20

25

30

Figure 11.11: After 2000 time steps, 73 molecules have stuck to the cold wall in the coldtrap and 27 are still “free”.


−1

0

1

2

3

4

5

6

7

8

5 10 15 20 25 30

5

10

15

20

25

30

Figure 11.12: After about 8000 time-steps all the molecules have stuck to the cold wall ofthe cold trap.

11.10. PROJECTS 353

You must write a MATLAB script M-file that does the following:

(a) Prompt the user for:

i. Chamber height and width

ii. Connecting pipe height and width.The pipe height must be less than or equal to the chamber height and greaterthan 0.

iii. The number of molecules to simulate.

(b) Display the initial geometry as in first figure.

(c) Place the molecules in the left chamber and display the initial configuration as inthe second figure.

(d) Execute the time steps until all the molecules stick to the right most wall. Displaythe final configuration as in fourth figure as well as the number of time-steps ittook to do the simulation.

Some hints:

• When testing your code, use geometries with just a few cells for height and widthand just a few molecules. This will allow the simulations to complete quickly andwill test the more subtle parts of the code, the wall collisions and the action ofsticking to the cold wall.

• When testing your code, make lots of use of visual output. Put in some graph-ical output for some or all of the time-steps with a pause statement followingthe graphics instruction. For example, the third figure shows some intermediategraphical output. This is a very valuable way of debugging the operation of yourcode.

• You may need the following MATLAB function in your code (it is not mentionedin the book!) to place the molecules randomly at the start. Using help uint32:

UINT32 Convert to unsigned 32-bit integer.

I = UINT32(X) converts the elements of the array X into unsigned

32-bit integers. X can be any numeric object (such as a DOUBLE). The

values of a UINT32 range from 0 to 4294967295. Values outside this

range are mapped to 0 or 4294967295. If X is already an unsigned

32-bit integer array, UINT32 has no effect.

.

.

.

See also ...UINT8, UINT16, INT8, INT16, INT32.


7. Solve the universe, plane and simple

Programmers are an ambitious lot, and so this problem is to simulate the evolution ofthe universe, assuming that you know the position and velocity of every object in it1.After solving this problem, anything else thrown at you in the rest of your career willseem like a “piece of cake”. For simplicity we will assume that the universe is only3-dimensional (two for space, one for time) instead of 4-dimensional and that eachobject in the universe is “point-like” with mass but with no spatial dimensions, and istravelling at a speed much less than the speed of light.

There are two parts to this problem, both to be solved with MATLAB. You cansolve them using any numerical/computational technique you like so long as you dothem with MATLAB. You must write and use a function M-file called timeStep that“evolves” the universe over a small increment of time DT. This function and the algo-rithms that one can use in it are described in some detail below. You will also write ascript M-file that will interrogate its user for the following inputs:

• Number of “objects” in your universe (greater or equal to 2),

• Mass of each object in kilograms,

• Starting x and y coordinates of each object in meters,

• Starting x and y velocities of each object in meters/second,

• Time alloted to each timestep, DT in seconds, and

• The total number of timesteps to devote to the evolution, NSteps.

Note that the magnitude of the velocity, sqrt(vx^2 + vy^2) must be much less thanthe speed of light, 3.0e8 m/s, for the equations described below to work sensibly. So,make sure that the input speed is less than 3.0e6. The user should be promptedrepeatedly for inputs until sensible inputs are provided.

The script M-file will also execute a loop over all the timesteps in the evolution, andprovide final graphical output. Examples of the program output will be provided below.

11.10.4 Supplementary material: Trajectories of objects inthe universe

The x and y components of acceleration (ai,x, ai,y) felt by the i’th object in the universeat position (xi, yi) because of gravitational attraction to the other N − 1 objects is:

axi =

N∑i�=j

−Gmj(xi − xj)

[(xi − xj)2 + (yi − yj)2]3/2

1Interestingly enough, at the dawn of the 20th century, theoretical physicists actually thought that wecould do this. Then, quantum mechanics came along and things got a lot more “uncertain”.

11.10. PROJECTS 355

ayi =

N∑i�=j

−Gmj(yi − yj)

[(xi − xj)2 + (yi − yj)2]3/2

where G is the gravitational constant (6.673 × 10−11 nt-m2/kg2). Note that the sum∑Ni�=j excludes the i’th object, as an object (being point-like) does not exert any grav-

itational force on itself.

11.10.5 Supplementary material: Stepping algorithms

If we assume that any object feels a constant acceleration over a small time interval, ∆t,then we may construct a stepping algorithm as described in the following pseudocode:

(a) Start with time t = 0.At t = 0 the starting conditions:xi(0), yi(0), v

xi (0), v

yi (0) for i = 1, 2, 3 · · ·N

are known.

(b) Calculate the acceleration at the start of the step for each particle in the system:

axi (t) =

N∑i�=j

−Gmj(xi(t)− xj(t))

[(xi(t)− xj(t))2 + (yi(t)− yj(t))2]3/2

ayi (t) =

N∑i�=j

−Gmj(yi(t)− yj(t))

[(xi(t)− xj(t))2 + (yi(t)− yj(t))2]3/2

(c) Estimate the positions and velocities at the end of the timestep assuming constantacceleration.

xi(t+∆t) = xi(t) + vx,i(t)∆t+1

2ax,i(t)(∆t)2

yi(t+∆t) = yi(t) + vy,i(t)∆t+1

2ay,i(t)(∆t)2

vx,i(t+∆t) = vx,i(t) + ax,i∆t

vy,i(t+∆t) = vx,i(t) + ay,i∆t

(d) Increment the time at the end of the timestep, t = t+∆t.

(e) If the total time to evolve the universe has been reached, stop. Otherwise go backto step 2.

There are a few subtleties you will have to think about.


• How do you make a reasonable choice for ∆t?

Basically, unless you are prepared to do a lot of extra math and physics to workthis out, you have to guess. Leave this up to the user. I’ll be happy with analgorithm that works as described above.

• What basic inputs should I start with?

You can dream up your own universe. However, here are some astronomical factsthat you can use to set the scale of things:

– G: 6.673× 10−11 nt-m2/kg2

– The Sun: Mass 1.989e30 kg

– The Earth: Mass 5.9736e24 kg, Earth-Sun orbit radius 149,600,000,000 m

– The Moon: Mass 7.35e22 kg, Moon-Earth orbit radius 384,400,000 m

Velocities? Work these out for yourself assuming that the heavier partner isstationary and the fact that the Earth goes around the Sun once per year andthe Moon goes around the Earth once per month. If you want the data on otherplanets, do a little research. I found the above numbers by surfing the web.

• Is my evolution realistic?

Only in the limit that ∆t → 0. You will find that the above equations, thelonger you iterate them will increasingly violate the physical laws of the conser-vation of energy and momentum. There appears to be no satisfactory solution tothis problem although there are better algorithms than the one presented above.Discussion of this would take us far beyond the scope of this course.

• How do I check that my algorithm is reasonable?

Here are some suggestions:

– Set G = 0 and verify that your objects travel in straight lines.

– The two-object case has an exact mathematical solution. You can use thisto check your results. Almost every introductory physics book discusses thissolution.

Here is an example of the Earth and Moon executing one orbit around the Sun. Inthis case, I used a timestep of 3600 seconds and 365.25× 24 timesteps.

11.10. PROJECTS 357

0 1 2 3 4 5 6 7 8 9 10

x 105

−5

−4

−3

−2

−1

0

1

2

3

4

5x 10

5

x−axis (m)

y−ax

is (

m)

Wobble of the Sun’s center of mass

Figure 11.13: This is the wobble of the center of mass of the Sun.


−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

x 1011

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2x 10

11

x−axis (m)

y−ax

is (

m)

Earth’s orbit

Figure 11.14: This is the Earth’s orbit.

11.10. PROJECTS 359

1.2 1.25 1.3 1.35 1.4 1.45 1.5 1.55 1.6 1.65

x 1011

2.5

3

3.5

4

4.5

5

5.5

6

6.5

7

x 1010

x−axis (m)

y−ax

is (

m)

Earth and moon together

Figure 11.15: This is part of the combined orbit of the Earth and the moon.


8. Re-Mastermind

You will write MATLAB code, a simpler version of the Mastermind program of Chapter8, one in which you, the user, try to guess the 4 hidden numbers.

You already know the rules of the game. Only, in the case, to make things simpler, theuncovered numbers are represented by ’0’s rather than ’?’s. (Mixing characters andnumbers in MATLAB is a bit of a pain.)

And, the game continues until you either quit, or win. Pretty simple, huh?

Hints:

How the computer finds the hidden numbers

answer = floor(9*rand(1,4) + 1);

Confused? Try:

>> help rand

>> help floor

How can you input 4 numbers at once?

guess = input(’Input your guess (1 <= [a,b,c,d] <= 9), [0,0,0,0] to quit: ’);

The above line prompts the user for input (See next page.) and the user can typein the elements of a row vector.

Confused? Try:

>> help input

An example play of the game is on the following page.

11.10. PROJECTS 361

>> MasterMind

Input your guess (1 <= [a,b,c,d] <= 9), [0,0,0,0] to quit: [1,1,1,1]

uncovered =

0 0 0 0

board =

1 1 1 1


uncovered =

0 0 2 0

board =

1 1 1 1

2 2 2 2


uncovered =

0 0 2 0

board =

1 1 1 1

2 2 2 2

3 3 2 3


uncovered =

0 0 2 0

board =

1 1 1 1

2 2 2 2

3 3 2 3

4 4 2 4


uncovered =

0 5 2 0

board =

1 1 1 1

2 2 2 2

3 3 2 3

4 4 2 4

5 5 2 5


uncovered =

0 5 2 6

board =

1 1 1 1

2 2 2 2

3 3 2 3

4 4 2 4

5 5 2 5

6 5 2 6


uncovered =

7 5 2 6

board =

1 1 1 1

2 2 2 2

3 3 2 3

4 4 2 4

5 5 2 5

6 5 2 6

7 5 2 6

ans =

You win!


9. Can computers cure cancer?

No, they can’t. But radiation can kill cancer and computers help. The followingproject outlines the essential problems of killing cancer with radiation. But first...

Some backgroundThere is about a 50% chance that you will get cancer. You almost certainly knowsomeone in your immediate family who died of cancer. My father died from it 3 yearsago. My mother and sister have had it, and, I will most certainly get it if I live longenough. Here are some startling facts:

• In the US alone, there are 1.4 million new cancer patients each year.

• Half of these will get treated with radiation. Cancer treatment using radiation iscalled radiotherapy.

• About 20% of those who are treated will die unnecessarily, even though cancerwas well localized. The radiation treatments were not effective enough to curethem. If you do the math, that’s 140,000 people per year in the US, about 50%more people than live in Ann Arbor!

• So, if you can make radiotherapy just 1% better, you will save 1,400 people peryear.

That’s what my research is all about—making radiotherapy better, using computersand computer algorithms. I thought I would give you a flavor of what the mainproblems are. And, I want to give you a MATLAB project a little more meaningfulthan Mastermind!

Hints and demonstrations with be given in class.

Here’s what you have to do.

(a) Make a model of a cross section of the human body 30×40 cm that is representedby a 30 row by 40 column array called body. This array will represent the energydeposited in the body by the radiation beams.Rows = 30; Cols = 40; body = zeros(Rows, Cols);

(b) Make a working array that pads an extra row or column on the exterior of body.This accounts for the outside. A(2:Rows + 1,2:Cols + 1) = body;

A(1,:) = 0; A(:,1) = 0; A(Rows + 2, :) = 0; A(:,Cols + 2) = 0;

(c) The tumor is a 4×4 cm square in the center of the body. Use pcolor and plot

to show the body and the outline of the tumor before the radiation is present.Here’s a hint:

show = A(2:Rows + 2,2:Cols + 2); %Handles off by one quirk in pcolor

%Some code to scale show so that it’s maximum value is 1.

%Hint: help max

11.10. PROJECTS 363

hold on

pcolor(0:Cols,0:Rows,show)

shading flat

plot(...you figure out the arguments...)

It should look like this2...

0 5 10 15 20 25 30 35 400

5

10

15

20

25

30

Figure 11.16: Before the radiation

(d) Give the user of the program the capability of putting up to 4 beams on thepatient from the outside, one from the top, one from the bottom, one from theleft and one from the right. For example...

>> TreatCancer

Place beam top to bottom? (0 = no, 1 = yes) 1

Place beam bottom to top? (0 = no, 1 = yes) 1

Place beam left to right? (0 = no, 1 = yes) 1

Place beam right to left? (0 = no, 1 = yes) 1

which should look like this...

2This is much prettier in color!


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Figure 11.17: After beams are placed.

11.10. PROJECTS 365

The beams decay (or attenuate) exponentially from the surface with a decayconstant of 15 cm. So, if a surface tile has energy energy value 1, at depth i,where i represents the depth in centimeters, the value should be exp(-i/15).Note that the beams should cover the entire extent of the tumor.

(e) If that was the whole story, cancer would be beaten! Unfortunately, radiationspreads, the tumor does not get a full dose and the healthy tissue can get damaged.A simple model of this spreading is to repeatedly average the energy value asfollows:

Ai,j = (Ai−1,j +Ai+1,j +Ai,j−1 +Ai,j+1)/4

Perform this averaging 5 times (perform the operation on the array and thenrepeat the process 4 more times) to produce a result that looks like the followingfigure.

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Figure 11.18: What happens after the beams spread.

Open ended questions. (Optional.) Solve these and collect numerous Nobel prizes formedicine and physics!

• Model a real 3-dimensional person rather than the simple geometry above.


• Can you optimize the treatment by using more beams? (The answer is yes. Thedifficulty is in finding the optimum configuration.)

• What is the best decay constant to use? (This is related to the energy of thebeam.)

• Develop the radiation physics to make the above calculation resemble reality.

• Design a better radiation machine that will deliver better beams for radiotherapy.

• Get this calculation done in 5 minutes or less!

• Find out the real reason why radiation kills cells.

• Find out what some radiation is good for you (far in excess of natural radiationexposure). This is called radiation hormesis.

• What role does radiation damage play in the development of cancers. This iscalled oncogenesis.

Chapter 12

Graphics

12.1 Two Dimensional Plots

12.1.1 A basic plot

% Note the use of pointwise operators

x = linspace(-pi,pi,1000);

y = (x.^2).*sin(20*x) + x;

plot(x,y)

pause % Use a "pause" statement to stop

close % Closes the figure

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12.1.2 Using different line colors

% Choose from:

% b: blue

% g: green

% r: red

% c: cyan

% m: magenta

% y: yellow

% k: black

% w: white

367

368 CHAPTER 12. GRAPHICS

plot(x,y,’b’) % blue

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plot(x,y,’g’) % green

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plot(x,y,’r’) % red

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plot(x,y,’c’) % cyan

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12.1. TWO DIMENSIONAL PLOTS 369

plot(x,y,’m’) % magenta

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plot(x,y,’y’) % yellow

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plot(x,y,’k’) % black

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plot(x,y,’w’) % white

% White!!!!!

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12.1.3 Making titles

% Titling your graph

plot(x,y,’w’) % white

title(’Plotting white on white is dumb!’)

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15Plotting white on white is dumb!

12.1.4 Axis labels

% Axis labels

plot(x,y,’k’)

xlabel(’This is the x-axis’);

ylabel(’\fontsize{16}y-axis’)

title(...

’\fontsize{20}\it Same boring plot!’)

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This is the x−axis

y−ax

is

Same boring plot!

12.1.5 Different line styles

% A different 2D example

u = linspace(-3, 3, 1000);

v = exp(-u.^2);

plot(u,v)

title(’The bell shaped curve’)

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% Plotting with different line styles

% Choose from:

% - solid line (default)

% : dotted line

% -. dash-dot line

% -- dashed line

%

% syntax plot(u,v,’:’)

% Note use of different colors and line styles

plot(u,v,’k:’)

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plot(u,v,’g-.’)

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plot(u,v,’r--’)

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plot(u,v,’b-’)

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% Example - The LogGamma function

s = linspace(0.01, 6, 20);

t = log(gamma(s));

plot(s,t)

title(...

’\fontsize{20}The log\Gamma(x) function’)

xlabel(’\fontsize{20}x’)

ylabel(’\fontsize{20}y’)

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The logΓ(x) function

x

y

12.1.6 Point plots, plotting with point symbols

% Plotting with point symbols

% Choose from:

% . point o circle

% x cross + plus sign

% * asterisk S square

% d diamond

% V triangle down ^ triangle up

% < triangle left > triangle right

% p pentagram h hexagram

%

% syntax plot(s,t,’:’)


% Note use of different colors and line styles

plot(s,t,’k:.’)

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plot(s,t,’b-+’)

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plot(s,t,’r--d’)

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plot(s,t,’g>’)

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plot(s,t,’kh’)

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12.1.7 Plotting more than one thing at once

% More arguments to the plot function

% Changing axes limits

clear all

x = linspace(-1, 1, 30);

f0 = ones(1,length(x));

f1 = x;

f2 = (1/2)*(3*x.^2 - 1);

f3 = (1/2)*(5*x.^3 - 3*x);

plot(x,f0,’k-’,x,f1,’k:’,x,f2,’k-.*’,x,f3,’k--o’)

title(’\fontsize{20}Plotting more than one thing on the same plot’)

xlabel(’\fontsize{20}x’)

ylabel(’\fontsize{20}y’)

axis([-1,1,-1,1.1]) %[xmin,xmax,ymin,ymax]

grid on % Turn the grid on

legend(’P0’,’P1’,’P2’,’P3’)

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Plotting more than one thing on the same plot

x

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P0P1P2P3


% Another way

hold on % Don’t erase when drawing new plot

plot(x,f0,’k-’)

plot(x,f1,’k:’)

plot(x,f2,’k-.*’)

plot(x,f3,’k--o’)


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% Another way - matrix form

g(1,1:length(x)) = f0;




plot(x,g)


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12.1.8 Subplots

% Yet another way, subplots

subplot(2,2,1), plot(x,f0,’k-’), title(’Plot 1’)

subplot(2,2,2), plot(x,f1,’k:’), title(’Plot 2’)

subplot(2,2,3), plot(x,f2,’k-.*’), title(’Plot 3’)

subplot(2,2,4), plot(x,f3,’k--o’), title(’Plot 4’)


zoom on

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12.2. THREE DIMENSIONAL GRAPHICS 379

12.2 Three dimensional graphics

12.2.1 Plot3 plots

% Example using plot3

clear all

t = linspace(0, 10*pi, 1000);

plot3(sin(t),cos(t),t)

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12.2.2 Mesh plots

% Mesh plots

[x,y,z] = sphere(30);

mesh(x,y,z)

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12.2.3 Axis labelling

% Provide a label for the x-axis

xlabel(’\fontsize{20}x-axis’)

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% Provide a label for the y-axis

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% Provide a title

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12.2.4 meshc and meshz plots

% Two other forms of mesh plots

% meshc and meshz

[x,y,z] = peaks(30);

subplot(1,2,1)

meshc(x,y,z),title(’\fontsize{20}meshc’)

subplot(1,2,2)

meshz(x,y,z),title(’\fontsize{20}meshz’)

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12.2.5 Surface plots using surf


surf(x,y,z)

title(’\fontsize{20}Example using surf’)

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shading flat % Flat shading

title(’\fontsize{20}Using shading flat’)

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shading interp % Interpolated shading

title(...

’\fontsize{20}Using shading interp’)

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12.2.6 Surface plots using surfc

surfc(x,y,z) % surf with contours

title(’\fontsize{20}Example using surfc’)

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Example using surfc

12.2.7 Surface plots using surfl

surfl(x,y,z) % surf with lighting

shading interp

colormap gray

title(’\fontsize{20}Example using surfl’)

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12.2.8 Contour plots using contour


contour(x,y,z,20)

title(’\fontsize{20}Example using contour’)

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colorbar % Show the colorbar

title(’\fontsize{20}With shading interp’)

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12.2.9 Contour plots using contour3

contour3(x,y,z,20)

title(’\fontsize{20}Example using contour3’)

grid off

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12.2.10 Contour plots using pcolor

% Example using pcolor

pcolor(x,y,z)


title(’\fontsize{20}Example using pcolor’)

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shading interp % Interpolated shading

title(’\fontsize{20}with shading interp’)

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12.2.11 Contour plots using contourf

% Example using filled contour

contourf(x,y,z,12)


title(’\fontsize{20}Example using contourf’)

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12.2.12 Contour plots with labels

% Example using contour with labels

C = contour(x,y,z,12);

clabel(C)

title(...

’\fontsize{20}Example using contour with labels’)

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Chapter 13

Miscellaneous topics

13.1 Pitfall review

13.1.1 Review of for loops

This module gives some review on MATLAB’s implementation of for loops mostly by wayof example.

The general syntax of a for loop is

for k = vectorOrColumnList

% MATLAB statements

end

There are several ways of stepping through the elements of a row vector.

Example (Need vector value but not index):

The most common usage is C++-like syntax

%File: fex1.m

% For loop requiring vector value but not the index

for k = 1:3

fprintf(’k = %g\n’,k)

end

% Prints

k = 1

k = 2

k = 3

387


or

%File: fex1a.m


for k = 1:2:8

fprintf(’k = %g\n’,k)

end

% Prints

k = 1

k = 3

k = 5

k = 7

or

%File: fex1b.m


for x = 1.0:-0.1:0.0

fprintf(’x = %g\n’,x)

end

% Prints

x = 1

x = 0.9

x = 0.8

x = 0.7

x = 0.6

x = 0.5

x = 0.4

x = 0.3

x = 0.2

x = 0.1

x = 0

or

%File: fex1c.m


theta = linspace(0,90,3);

for thetai = theta

fprintf(’theta = %g\n’,thetai)

end

% Prints

13.1. PITFALL REVIEW 389

theta = 0

theta = 45

theta = 90

Example (Need vector value and index):

%File: fex2.m

% For loop requiring vector value and the index

v = linspace(0,250,3);

for i = 1:length(v)

fprintf(’velocity(%g) = %g\n’,i,v(i))

end

% Prints

velocity(1) = 0

velocity(2) = 125

velocity(3) = 250

Example (Nested for loops):

%File: fex3.m

% Nested for loops

theta = linspace(0,90,2);

v = linspace(0,250,2);

for thetai = theta

fprintf(’theta = %g\n’,thetai)

for i = 1:length(v)

fprintf(’velocity(%g) = %g\n’,i,v(i))

end

end

% Prints

theta = 0

velocity(1) = 0

velocity(2) = 250

theta = 90

velocity(1) = 0

velocity(2) = 250

Example (when the index is a column vector):

%File: fex4.m

% For loop where the index is a column vector


a = [1 2 3

4 5 6];

[rows,columns] = size(a);

fprintf(’a is a %g x %g matrix\n’,rows,columns)

for b = a

disp(b);

disp(’-------’)

end

% Prints

a is a 2 x 3 matrix

1

4

-------

2

5

-------

3

6

-------

Note in this example that b is a new column vector on each iteration of the loop.

13.1.2 Review of functions

Some detail is given in Chapter 13 of Austin and Chancogne. However, the definitive ref-erence for this course is given in the Chapter 11 that was handed out and is on the web.This lecture gives some review on MATLAB’s implementation of functions mostly by wayof example.

Recall:

The general syntax of the function definition is to have a separate file called functionName.m

that has the following form:

function [out1,out2,...,outN] = functionName(in1,in2,...inM)

%FUNCTIONNAME A brief one line description (optional)

% .

% .

% More description (optional)

% .

% .

% .


% First executable statement follows this line

.

.

.

% Valid executable MATLAB statements and comments

.

.

.

% Last line in functionName, control reverts to calling M-file or command window

The function call is made with the following statement:

[out1,out2,...,outN] = functionName(in1,in2,...inM)

Functions with no input or output arguments

Here’s a function called noEvil.m that does nothing except printing something silly:

function noEvil

%NOEVIL noEvil is a function that does nothing

%

% ENG101 class example

%File: noEvil.m

disp(’See no evil’)

disp(’Hear no evil’)

disp(’Do no evil’)

disp(’A function that does absolutely nothing’)

Functions with one output argument

Here’s a function called square.m, that computes the square of its argument:

function a2 = square(a)

%SQUARE square is a function that squares an argument

%

% It can accept scalars, vectors or matrices

%


%File: square.m

a2 = a.^2;


and some examples of its use on a scalar, a vector, a matrix and the element of a matrix:

>> square(2)

ans = 4

>> y = linspace(0,10,11)

y = 0 1 2 3 4 5 6 7 8 9 10

>> square(y)

ans = 0 1 4 9 16 25 36 49 64 81 100

>> x = round(10*rand(4))

x = 5 10 3 2

2 7 4 8

6 4 9 6

3 7 7 1

>> square(x)

ans = 25 100 9 4

4 49 16 64

36 16 81 36

9 49 49 1

>> x(3,3) = square(x(3,3))

x = 5 10 3 2

2 7 4 8

6 4 81 6

3 7 7 1

>> z = square(x(2,2))

z = 49

Functions with more than one output argument

Here’s a function called squareCube.m, that computes the square and cube of its argument:

function [a2,a3] = squareCube(a)

%SQUARECUBE squareCube is a function that squares and cubes its argument

%

%


%



%File: squareCube.m

a2 = a.^2;

a3 = a.^3;

and some examples of its use on a scalar, a vector and a matrix:

>> squareCube(2)

ans = 4

>> [u,v] = squareCube(2)

u = 4

v = 8

>> y = linspace(0,10,11)

y = 0 1 2 3 4 5 6 7 8 9 10

>> squareCube(y)

ans = 0 1 4 9 16 25 36 49 64 81 100

>> [u,v] = squareCube(y)

u = 0 1 4 9 16 25 36 49 64 81 100

v = 0 1 8 27 64 125 216 343 512 729 1000

>> x = round(10*rand(4))

x = 10 9 8 9

2 8 4 7

6 5 6 2

5 0 8 4

>> squareCube(x)

ans = 100 81 64 81

4 64 16 49

36 25 36 4

25 0 64 16

>> [u,v] = squareCube(x)

u = 100 81 64 81

4 64 16 49

36 25 36 4

25 0 64 16

v = 1000 729 512 729

8 512 64 343

216 125 216 8

125 0 512 64

>> [a,b] = squareCube(x(2,2))


a = 64

b = 512

Note that if you want to employ all the outputs, you have to assign them, else only thefirst output is displayed. These functions work on scalars, vectors, matrices and elements ofmatrices.

Subordinate functions

Let’s rewrite squareCube.m in the following way:

function [a2,a3] = squareCallsCube(a)

%SQUARECALLSCUBE squareCallsCube is a function that .^2’s and .^3’s its argument

%

% squareCallsCube calls a subordinate function called cube.m


%


%File: squareCallsCube.m

a2 = a.^2;

a3 = cube(a);

return

function c = cube(b)

c = b.^3;

squareCallsCube.m works the same way as if you had called squareCube.m. However, thefunction cube.m is only known to the function squareCallsCube.m.

>> cube(a)

??? Undefined function or variable ’cube’.

>> help cube

cube.m not found.

>>

13.2. THE MATLAB WAY 395

13.2 The MATLAB way

One of the most important things to remember is that MATLAB is organized around thevariable type of two-dimensional arrays of doubles. Hence, MATLAB’s intrinsic functionsare optimized to use this variable type. If you are making use of this aspect of MATLAB,you are doing things “the MATLAB way”.

If you are not (and could be), then your application will almost certainly run slower.

For example, let’s consider a temperature diffusion problem. (Examples of a heat diffu-sion calculation in 1 dimension is given in the example codes ironbarPointJacobi.m andironbarGaussSeidel.m. A 2-dimensional example may be found in the example codesheat.m, heatSlow.m, and heatFast.m) The part of the algorithm that takes all the time is:

for i = 2:N-1

for j = 2:N-1

T(i,j) = 0.25*(oldT(i-1,j) + oldT(i+1,j) + oldT(i,j-1) + oldT(i,j+1));

end

end

Here is a faster solution obtained by using the intrinsic matrix operations of MATLAB. Theessential change is made by converting the for loops to

and changing them to

T(2:N-1,2:N-1) = 0.25* ...

( ...

oldT(1:N-2,2:N-1) + oldT(3:N,2:N-1) + oldT(2:N-1,1:N-2) + oldT(2:N-1,3:N) ...

);

This will be demonstrated in lecture on a related problem. The speedup is highly dependenton computer architecture. On the laptop the typical speedup is a factor of only about 2 fora 50 by 50 array of temperature. On a high-end workstation, I’ve obtained as high as 30times faster.

13.3 Selected Applications

13.3.1 Free fall with air resistance

Let’s consider another trajectory problem that we will solve using the stepping method.Consider taking an object at some height, y0 and dropping it with zero velocity, v0 = 0.


This is a one-dimensional problem and can be solved completely with equations. However,we will execute a computer solution as a model for solving more complex problems, liketrajectories of baseballs. The air resistance on a baseball is very significant. If there were noair resistance, the longest home runs (about 600 feet or so) would travel half a mile!

The acceleration on the object is:a = Dv2 −G (13.1)

where D is the drag and G is the acceleration due to gravity.

We can “solve” this in a formal fashion as follows:

y(t) = y0 + v0t+∫ t

0dt′∫ t′

0dt′′ a[v(t′′)] (13.2)

In fact, even if a were dependent on position y and time t we could write a formal solutionas:

y(t) = y0 + v0t+∫ t

0dt′∫ t′

0dt′′ a[y(t′′), v(t′′), t′′] (13.3)

But, this is a formal solution only and does not lead to any practical answers, only methodsof attack on this problem.

One such attack on the problem is to divide the integral above over a set of time interval ∆tand set up an iterative equation as follows.

yi = yi−1 + vi−1∆t+∫ ti−1+∆t

ti−1

dt′∫ t′

ti−1

dt′′ a[y(t′′), v(t′′), t′′] (13.4)

which is still exact. However, if we make the approximation that the acceleration is constantover the step, that is

ai[y(t′′), v(t′′), t′′] ≈ ai[y(ti−1), v(ti−1), ti−1] (13.5)

we get

yi ≈ yi−1 + vi−1∆t+1

2ai[y(ti−1), v(ti−1), ti−1]∆t2 (13.6)

and an algorithm that we can program:

1. Set initial conditions y0, v0 at t = 0.

2. Initialize a counter i = 1.

3. Increment the counter i = i+ 1

4. Calculate the acceleration ai[y(ti−1), v(ti−1), ti−1]

5. Take a small step yi = yi−1 + vi−1∆t+ 12ai[y(ti−1), v(ti−1), ti−1]∆t2

13.3. SELECTED APPLICATIONS 397

6. Calculate the new velocity vi = ai[y(ti−1), v(ti−1), ti−1]∆t

7. Update the time ti = ti−1 +∆t

8. Go back to step 3 unless some stopping criterion is reached.

Note that two-dimensional problems are not really more complicated. There would just bemore variables to carry around.

Now the central question arises.How small should I make ∆t?????The answer to THIS question is the art of computer programming and lies at the heart of thegeneral field of “Numerical Methods”. The field of “Numerical Methods” is also concernedwith developing better and faster techniques for solving problems than by the “brute force”method outlined above. However, for this problem, we can more or less guarantee that ourresults are good if the acceleration changes very little over a time step, namely

da

dt∆t � a (13.7)

At this point, let’s simplify the discussion and consider just our problem. If we apply theabove condition to the problem we get

∆t � 1

2Dv(13.8)

Does this make sense? It does! If there is no drag, D = 0, and we can make ∆t anything welike! This is because with no drag there is constant acceleration and we don’t have to makethe steps small to get an accurate solution! (We have to be careful that we do not violateour stopping criterion, however. More on this later.)

What if the v is small? In this case we can set ∆t to be large because the drag is not affectingthe motion very much. If D and/or v are large enough such that Dv is large, we had betteruse small time steps because drag forces are strongly affecting the trajectory.

How do we know how to set ∆t for our problem? The greater the velocity, the greater carewe have to take by setting smaller time steps. The greatest velocity we can reach with ourproblem is called the “terminal” velocity. This is when the drag force and the accelerationforce exactly balance each other and there is no net acceleration. For free fall with airresistance, this velocity is

vT =

√G

D(13.9)

For our baseball the terminal velocity is something like 129 ft/s.


This is the upper limit of the velocity and so we are conservative by saying that

∆t � 1

2√DG

(13.10)

or ∆t � 2.

How much smaller than 2 should you make ∆t? Usually, as small as we can get away with!If you want the error to be of the order of 1% or less, you should set it to something below0.02 seconds. At least we have some indication from a simple analysis.

Now we implement the algorithm with MATLAB coding. Here is what the essential workingpart of the algorithm would look like:

y = 1000; % Starts at 1000 feet

v = 0 ; % with 0 velocity

while (y > 0)

a = D*v^2 - G; % Acceleration at beginning of step

y = y + v*dT + 0.5*a*dT^2; % Position at end of the step

v = v + a*dT; % Velocity at end of the step

end

Here’s what the complete code would look like. It gives some example of organization andplotting:

%File: drag.m

clear % Clears variables

echo off % Turns off echo

clc % Clears the screen

G = 32.174; % Force due to gravity

D = 0.0019285; % Drag on a baseball

dT = 10; % Initial time step

vT = sqrt(G/D); % Terminal velocity

fprintf(’Terminal velocity is %g ft/s\n’,vT)

while dT > 0

dT = input(’What delta time to use (0 or less to stop)? >’);

fprintf(’ Using dT = %g\n’,dT)

if dT <= 0

return % dT <= 0 used as an input sentinel


end



index = 1; % Index into the counting arrays

Y = y; V = v; T = 0; % Plotting arrays

while (y > 0)


y = y + v*dT + 0.5*a*dT^2; % Position at end of the step


% Plotting arrays

index = index + 1;

Y(index) = y;

V(index) = v;

T(index) = T(index - 1) + dT;

end

subplot(1,2,1)

zoom on

hold on

plot(T,Y,’-’)

title(’y as a function of t’)

xlabel(’time (s)’)

ylabel(’height (ft)’)

subplot(1,2,2)

zoom on

hold on

plot(T,V,’-’)

title(’v as a function of t’)

xlabel(’time (s)’)

ylabel(’height (ft/s)’)

fprintf(’Final velocity = %g\n’,V(end));

end

When we run this code, as demonstrated in class, we see that the trajectory always “over-


shoots”. We always end up with a negative y at the end. We have to make ∆t very small toget the final position close to zero. We can fix this problem relatively easily.

What we can do is check to see if the ∆t we are about to use will cause an overshoot and ifit will, to truncate it accordingly.

How do we truncate it? Recall that we are assuming that the acceleration over the courseof the step is constant. If that is so, we can calculate the ∆t required to bring it exactly tozero. It must satisfy the quadratic equation:

0 = yi−1 + vi−1∆t+1

2ai[y(ti−1), v(ti−1), ti−1]∆t2 (13.11)

You can solve this with the quadratic equation and convince yourselves that the solution is:

dT = -(v + sqrt(v^2 - 2*y*a))/a;

Here is the new working part of the algorithm incorporating this idea. (See the examplecode drag0.m.)



while (y > 0)


yTest = y + v*dT + 0.5*a*dT^2; % Position at end of the step

if yTest < 0

% Handle the overshoot

dT = -(v + sqrt(v^2 - 2*y*a))/a;

y = 0;

else

y = yTest;

end


end

So now we have a trajectory that always ends in the right place and an algorithm that isguaranteed to get us close to the right answer. We are now a long way towards solving morecomplicated problems, like two-dimensional trajectories of baseballs.

13.3.2 Diffusion of charged ions in an electric field

This is the problem of positive and negative ions created somehow (like a pulse of radiation)in a gas drifting in opposite directions under the influence of a constant electric field. Run


the following code called ion.m that resides in the classcodes area. Note how the leadingedge is more populated with ions than the trailing edges since the longer an ion spends inthe vicinity of ions of opposite charge the greater the chance that it will “recombine’ andneutralize.

%File: ion.m

clear all

echo off

clc

NSteps = 100;

NIon = 2000;

for i = 1:NIon

x1(i) = rand; y1(i) = rand;

x2(i) = rand; y2(i) = rand;

end

plot(x1,y1,’b*’,x2,y2,’r*’)

axis([-1 2 0 1])

axis off

title(’\fontsize{20}Initial configuration’)

pause

for time = 1:NSteps + 3*sqrt(NSteps)

for i = 1:NIon

if (rand < 2/NSteps & x1(i) < 1)

y1(i) = -100;

end

theta = 2*pi*rand;

x1(i) = x1(i) + (1 + cos(theta))*0.5/NSteps;

y1(i) = y1(i) + sin(theta)*0.5/NSteps;

if (rand < 2/NSteps & x2(i) > 0)

y2(i) = -100;

end

theta = 2*pi*rand;

x2(i) = x2(i) - (1 - cos(theta))*0.5/NSteps;

y2(i) = y2(i) + sin(theta)*0.5/NSteps;

end


plot(x1,y1,’b*’,x2,y2,’r*’)

axis([-1 2 0 1])

axis off

title(’\fontsize{20}\leftarrow +ions | -ions\rightarrow’)

pause(0.01)

end

13.3.3 Calculation of electric fields

Constant electric fields as employed in the problem above are really fictional. In this sim-ulation we mock-up a realistic electric field situation. Run the following code called phi.m

that resides in the classcodes area. Note how strong the electric field is near the insulator.(It helps to have come to class to get the full description of this problem!)

%File: phi.m

clear all

echo off

clc

N = input(’Number of cells = N x N. Input N: ’)

delta = input(’Convergence criterion delta: ’)

V = zeros(N);

V(1,N/2:N) = 1000;

V(N,N/2:N) = 1000;

V(:,N) = 1000;

oldV = V;

change = 10;

tic

counter = 0;

while(change > delta)

V(2:N-1,2:N-1) = ...

0.25*( ...

V(1:N-2,2:N-1) + V(3:N,2:N-1) + V(2:N-1,1:N-2) + V(2:N-1,3:N) ...

);

difference = abs(V - oldV);



counter = counter + 1;

if mod(counter,100) == 0

fprintf(’Interation %g, average change = %g\n’,counter,change)

end

oldV = V;

end

toc

pcolor(1:N,1:N,V);

shading interp;

xlabel(’\fontsize{20}X-grid’);

ylabel(’\fontsize{20}Y-grid’);

title(’\fontsize{20}Voltage in the chamber’);

pause

figure

[Ex,Ey] = gradient(V);

E = sqrt(Ex.^2 + Ey.^2);

warning off

pcolor(1:N,1:N,log(E));

shading interp;



title(...

’\fontsize{20}Mag. of electric field in the chamber’);

pause

figure

surf(1:N,1:N,log(E));

shading interp;



title(...

’\fontsize{20}Mag. of electric field in the chamber’);


13.4 Summary of the Course

By now you are conversant in two languages, C++ and MATLAB. You have noted thedifferences between them. C++ is hard to make work properly. It is a powerful, general-purpose language that runs numerical calculations very fast.

MATLAB is specialized for engineering-type problems. It has lots of engineering-type utilitybuilt in and has very nice plotting capabilities, particularly in 3 dimensions. It is easy to getan application “up and running”, but it is relatively slow for large-scale problems.

Not every computer language is suitable for every purpose. Here’s how a research scientistwould use both of these computer languages to his or her best advantage.

1. Design a computer algorithm in pseudo-code

2. Convert it to MATLAB

3. Debug the concept of the algorithm using MATLAB

4. Develop graphics output to interpret results

5. Develop a set of test problems that can be completed by MATLAB in a reasonabletime.

6. When the MATLAB version is acceptable, re-code the algorithm in C++ starting withthe MATLAB code.

7. Run the same set of test problems using the C++ code.

8. Make sure that the C++ results and the MATLAB results of test problems are iden-tical.

9. When the C++ code passes the tests, increase array size or introduce sufficient com-plexity to solve the problem you really want to solve.

10. Have C++ write data arrays that MATLAB can read in an display, using MATLAB’sexcellent graphics capabilities.

11. Publish results!

The upshot is that both ways of doing it has its strengths and weaknesses. Why not useboth?

Chapter 14

Programming Style Guide for C++

This programming style guide is a set of mandatory requirements for C++ code layout. Thesestylistic “laws” are not made for arbitrary reasons. Experience teaches us that conforming tothis style, more or less standard through the industry with few exceptions, makes code easyto maintain and allows it to be read less painfully by others. This style guide is mandatoryfor ENG101, and can only be relaxed under specific instructions from GSIs and Profs.

This style guide is summarized by example:

• Indent 3 spaces relative to the preceding for every new nesting level. Curly bracketsmust always align vertically. Example,

int main(void)

{

int i = 0, k = 1;

if (i < k)

{ int temp = i;

i = k;

k = temp;

if (0 == k)

{ i = i*i;

k = -i;

}

}

return 0;

}

• Variable and function names are in lowercase, terse but descriptive, enough informationconveyed within the programming context, with capital letters to start new wordsinternal to the identifier. Example,

405

406 CHAPTER 14. PROGRAMMING STYLE GUIDE FOR C++

numStuds = 211;

degToRad = 180/PI;

Don’t over do it! Bad example,

numberOfStudentsInTheENG101Setion200Class = 211;

conversionFactorForDegreesToRadians = 180/PI;

The first encounter with a variable merits a short comment unless the meaning iscompletely obvious. Example,

numStuds = 211; //Number of students in the ENG101 Section 200 class

degToRad = 180/PI; //Conversion factor for degrees to radians.

• Never use underscores in function and variable names. Bad example,

this_variable_name_is_difficult_to_type = 211;

cute_______________but_dumb = true;

• Constant names should be in uppercase. Leading underscores are forbidden. Under-scores in the middle are acceptable although discouraged. Example,

const float PI = 3.14159;

const float EULER_CONSTANT = 0.577216;

• A variable should be initialized near where it is first used and should only be definedwithin the scope for which it is relevant. Examples,


float x[SIZE], y[SIZE];

for (int i = 0; i < SIZE; i = i + 1) // i only needed with scope of for loop

{ x[i] = 0;

y[i] = 0;

}

float z[SIZE] = {0};

int i; // i needed outside scope of for loop

for (i = 0; i < rand()%SIZE; i = i + 1) z[i] = 1;

cout << "The first " <, >=, &&and ||. Examples,

407

float result = x + y - z; //Good

float result=x+y-z; //Bad

• There are no spaces before and after the binary operators *, /, and %. There is nospace following the unary operators - and !. Examples,

float result = x*y + z; //Good

bool decision = ! go; //Bad

• No hardwired constants!

Bad example:

int a[10][10];

for (i = 0; i == 9; i = i + 1)

for (j = 0; j == 9; j = j + 1)

a[i][j] = 1;

Good example:


int a[SIZE][SIZE];

for (i = 0; i < SIZE; i = i + 1)

for (j = 0; j < SIZE; j = j + 1)

a[i][j] = 1;

• Functions should be small enough so that they fit on the screen. That’s about 30 linesof code.

• The use of goto, break and continue is disallowed.

• Use of non-constant global variables is discouraged.

C++ Code layout

Every C++ file, including main() is laid out as follows:

• Header comment block

• #include statements

• Constants


• Global variables

• Functions, with own headers. Function prototypes are optional unless required byordering constraints.

Here is an example of code layout:

409

// HEADER COMMENTS GO AT THE TOP ************************************

/********************************************************************

Copyright (C): University of Michigan

Project: ENG101/200 - Class example

File: C++example.cpp

Purpose: Coding example to show code layout

Compiler: g++

Programmer: Alex Bielajew


*********************************************************************/

// INCLUDE statements ***********************************************

#include <iostream>


// CONSTANTS ********************************************************

// No constants in this example

// GLOBAL VARIABLES *************************************************

// No global variables in this example

// FUNCTIONS ********************************************************

float pow(float x, int n)

/*********************************************************************

Purpose: Calculate the integral power of a float using the algorithm

of al-Kashi

Receives: "x" which is to be raised to the power "n"

Returns: a float, the result of x raised to the power n

*********************************************************************/

{ float result; // The result of the calculation


if (0 > n)

{ n = -n;


x = 1/x;

}

result = 1;

while(1 <= n)

{ if (0 == n % 2)

{ n = n/2;

x = x*x;

}

else

{ n = n - 1;

result = result*x;

}

}

return result;

}

411

int main(void)

/*********************************************************************

Purpose: To prompt the user for a float x and and int n and call a

function that computes x to the power of n

*********************************************************************/

{ cout << "A calculation of x (float) to the power of n (int) \n";

float x; // The float to be raised to a power

int n; // Power to be raised to

cout << "Input x (float) and n (int): ";

cin >> x >> n;


result = pow(x,n);

cout << x << " to the power " << n << " = " << result << "\n";

return(0);

}


Chapter 15

Syntax reference for beginning C++

4 Bit Computer Arithmetic

BINARY HEX UNSIGNED SIGNED BINARY ARITHMETIC (BASE 2) i (integer), ~i (bit-complement of the integer)

0000 0 0 0 0000 + ? = ? + 0000 = ? The integer 5 in 4 bit binary is 0101

0001 1 1 1 0001 + 0001 = 0010 The complement of 5 is ~5 = 1010

0010 2 2 2 0010 + 0001 = 0001 + 0010 = 0011 5 + ~5 = 0101 + 1010 = 1111

0011 3 3 3 0011 + 0001 = 0100 5 + ~5 + 1 = 1111 + 0001 = 0000

0100 4 4 4 0010 + 0010 = 0100 -i = ~i (bit-compliment of i) + 1

0101 5 5 5 0001 * ? = ? * 0001 = ? -5 = ~5 + 1

0110 6 6 6 0010 * 0011 = 0110 1011 = 1010 + 0001

0111 7 7 7 0011 * 0011 = 1001

1000 8 8 -8

1001 9 9 -7

1010 A 10 -6

1011 B 11 -5

1100 C 12 -4

1101 D 13 -3

1110 E 14 -2

1111 F 15 -1

32 Bit Computer Arithmetic

BINARY HEX UNSIGNED SIGNED i = 00000000000000000000000000010001

00000000000000000000000000000000 00000000 0 0 ~i = 11111111111111111111111111101110

00000000000000000000000000000001 00000001 1 1

00000000000000000000000000000010 00000002 2 2 i 00000000000000000000000000010001

... ... ... ... + ~i = + 11111111111111111111111111101110

00000000000000000000000000001111 0000000F 15 15 ---- ----------------------------------

00000000000000000000000000010000 00000010 16 16 11111111111111111111111111111111

00000000000000000000000000010001 00000011 17 17

... ... ... ...

01111111111111111111111111111111 7FFFFFFF 2,147,483,647 2,147,483,647

10000000000000000000000000000000 80000000 2,147,483,648 -2,147,483,648

... ... ... ...

11111111111111111111111111111110 FFFFFFFE 4,294,967,294 -2

11111111111111111111111111111111 FFFFFFFF 4,294,967,295 -1

413

414 CHAPTER 15. SYNTAX REFERENCE FOR BEGINNING C++

Algorithms and Pseudocodes

An Algorithm is a set of simple (i.e. well-defined) instructions to be carried out in sequence to accomplish some task. It

has a start and a stop and can execute branches and loops.

A Pseudocode is a way of expressing the actions and ordering to be taken place on a computer in an exact yet informalway, without details or syntax of a particular computer language.

e.g. IF such-and-such THEN DOthis-and-that

OTHERWISE DOsomething-else

Order of Precedence Logical Expressions

OPERATOR ASSOCIATIVITY

() left to right

* / % left to right

+ - left to right

< <= > >= left to right T && T = T, T && F = F, F && T = F, F && F = F

== != left to right T || T = T, T || F = T, F || T = T, F || F = F

&& left to right !T = F, !F = T

|| left to right False is the same as 0, True is anything else

= right to left

415

If/Else

if(logical expression 1) • else if’s are optional

{ STATEMENT BODY 1 • else is optional independent of the existence of else if’s

} • If there is only one statement in the statement body, you

else if(logical expression 2) may ignore the {}, but for clarity it is better to keep them

{ STATEMENT BODY 2

}

// more else if’s if needed

else

{ STATEMENT BODY N

}

Loops

While

while(logical expression) • If the logical expression is False the first

{ STATEMENT BODY time, the statement body is never executed

} • If the logical expression is always True, then

you have an infinite loop

Do/While

do • The statement body is evaluated once before

{ STATEMENT BODY the logical expression is evaluated at all

}while(logical expression); • Do NOT forget the ; at the end

For

for(expression 1; logical expression; expression2) • expression 1 is optional

{ STATEMENT BODY • logical expression is optional

} • expression 2 is optional

• ;; are not optional

• for(;;){} is an infinite loop

Example Programs

CODE OUTPUT CODE OUTPUT

#include<iostream> 1 #include<iostream> 1

416 CHAPTER 15. SYNTAX REFERENCE FOR BEGINNING C++

using namespace std; 2 using namespace std; 2

3 int main(void) 3

int main(void) 4 { int j = 0; 4

{ int i; 5 5

for(i = 1; i <= 10; i = i + 1) 6 do 6

cout <


int main(void) i is 2. j is -2.

{ int i = 2,j = -2; i is 1. j is -1.

while(j <= i)

{ if(j < 0 )

{ cout << "i is " << i <<". "

<< "j is " << j << ".\n";

i = i - 1;

}

j = j + 1;

}

return 0;

}

Chapter 16

Syntax reference for more advancedC++

Addresses

The address of a variable is its location in memory—typically some number from between 0 and 232 − 1.

The address of a variable is obtained with the unary & operator.

The piece of code {int i; cout << &i << "\n";} should be read as “i is declared an int. Print the address of i in hexadecimalformat.”.

Functions, call by value, call by reference

There are two important things to remember for functions: 1) the function definition and 2) the function call:

The function definition is written at the top of the C++ file outside of main or any other function. A function definitionmust be given before any code that uses the function via a function call. Generic syntax:

[void|int|float|...] functionName(variable type & name list separated by ","’s){C++ code}

Examples: int function1(int a, float b){int c;...;return c;}

void function2(int& c, float& d){...;return;}

Note: Variable types and names in parameters lists of function definitions MUST BE INCLUDED.If the return value is NOT void, the function must return the correct variable type.If the return value IS void, the return; is OPTIONAL.Functions can “return” at most ONE VARIABLE.

The function call is made inside of main or another function. Generic syntax:

[variable =] functionName(variable name list separated by ","’s);

Example: int a,c;float b;...;c = function1(a, b);

Note: Variable names in parameters lists of function calls MUST BE INCLUDED.Variable types in parameters lists of function calls MUST NOT BE INCLUDED.

417

418 CHAPTER 16. SYNTAX REFERENCE FOR MORE ADVANCED C++

Function call by value means that the parameter list of the function definition contains values. Example:

float function1(int c, float d){return d/c;} // function definition

x = function1(a, b); // function call

Note: Copies of these values are made available to the function. These are lost when the function returns.

Function call by reference means that the parameter list of the function definition contains addresses. Example:

function2(int& c, float& d){d = d/c;} // function definition

function2(a, b); // function call

Note: Functions called by reference can change the contents of the addresses that are made available to thefunction in the reference list.

VectorsUsing vectors

#include <vector> // Must be included in the pre-processor area

Note: Vectors are a “class” defined in standard C++

Declaring vector<int> vectorName(10); // An int vector called vectorName of size 10;

Note: Vectors are initialized to 0 when they are declared.

Accessing vector elements

for(i = 0; i < 10; i = i + 1){vectorName[i] = i + 1}; // for loop assigns all elements of vectorName

i = vectorName[0]; // i is assigned 1st element of vector vectorName

vectorName[9] = i; // The last (10th) element of vector vectorName is assigned the value of i

Note the “off-by-one” syntax in the above three examples

Passing vector to functions

double function1(vector<double> b){...} // Typical definition for call-by-value

vector<double> a(10);...;x = function1(a); // Typical call-by-value

double function2(vector<double>& d){...} // Typical definition for call-by-reference

vector<double> c(10);...;x = function2(c); // Typical call-by-reference

Vector class functions

vector<double> a(10);cout << a.size(); // Prints the size of the a vector

vector<float> b(10);b.pop_back(); // Reduces the size of the b vector by one to size 9

vector<int> c(10); int d = 42; c.push_back(d); // Increases the size of the c vector by one to size 11,

// with value of int d in the last vector element

Two dimensional (2D) arraysDeclaring int arrayName[2][3]; // Array called arrayName of size 2 rows by 3 columns

Declaring and initializing

int arrayName[2][3] = {{1,2,3},{4,5,6}}; // Initialize all the elements of arrayName;

int arrayName[2][3] = {{1,2,3}}; // Initialize the first row of arrayName, rest to zero;

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int arrayName[2][3] = {0}; // Initialize all the elements of arrayName to zero;

Accessing array elements

// for loops below assign all elements of arrayName

for(i = 0; i < 2; i = i + 1) for(j = 0; j < 3; j = j + 1) arrayName[i][j] = 3*i + j;

i = arrayName[0][2]; // i is assigned 1st row/3rd column element of array arrayName

arrayName[1][1] = i; // The last 2nd row/2nd column element of array arrayName is assigned the value of i

Note the “off-by-one” syntax in the above three examples

Note: arrayName is the address of arrayName[0][0], that is, arrayName is the same as &arrayName[0][0]

Passing 2D arrays to functions Remember that an array name is the address of the start of the array

// Note that column dimension must be specified below

void function(double b[][3], int rowSize, int columnSize){...} // Typical definition

double a[2][3];...;function(a); // Typical call

Note that that the column size (second index) must be known by the function!

Characters and stringschar’s are one-byte integers, strings are arrays of chars that includes a special ending character.

Declaring a character char ch; // Declares a character;

Declaring and initializing a character

char ch = ’a’; // Declares a character called ch and initializes it to ’a’. Note single quotes.

Declaring a string string name; // Declares a string

Declaring and initializing a character string

string name = "Albert Einstein"; // Declares a string

Note the use of double quotes in assigning strings.

Printing strings

string name = "Barry Manilow"; cout << name;

Structuresare compound data types. Structures have to be: 1) defined, 2) declared, 3) initialized, 4) used.

Defining a structure

struct AgeAndHeight {int age; float height;}; // Structure tag AgeAndHeight with an int and a float;

Declaring a struct

struct AgeAndHeight Wilma; // Declares Wilma as an AgeAndHeight struct

Declaring and initializing a character

struct AgeAndHeight Wilma = {26,65.5}; // Declares Wilma as an AgeAndHeight struct and initializes it

Referencing struct elements

420 CHAPTER 16. SYNTAX REFERENCE FOR MORE ADVANCED C++

Wilma.age = 27; // Wilma’s age is assigned the value 27

cout << "Wilma’s height is: " << Wilma.height << endl; // Prints Wilma’s height

Struct vectors

vector<struct AgeAndHeight> theFlintstones(4); // A vector of struct’s of the AgeAndHeight type

cout << theFlintstones[0].height; // Prints the height of 1st element

Struct arrays

struct AgeAndHeight townOfBedrock[100][5]; // A array of struct’s of the AgeAndHeight type

Struct vectors and arrays are passed to functions in the same way as the standard variable types.

RecursionFunctions can call themselves. Factorial example:int factorial(int n)

{ if (0 == n) return 1;

else return (n * factorial(n - 1));

}

The guidelines are:

• Successive calls to functions should represent progressively simpler cases. (For example, (N − 1)! is “simpler” than N !)

• There is an end to the recursion, the “base case”, that does not involve a recursive call. (For example 0! = 1) Not doing this

will lead to a program or computer crash.

• If a function calls itself more than once, and the “nesting level” gets deep, the calculation can be very inefficient.

Chapter 17

Syntax reference for MATLAB

MATLAB Workspace and Variables

The basic variable type in MATLAB is a two-dimensional array of doubles (64-bit represen-tation).A scalar is a 1× 1 array.A row vector of length N is a 1×N array.A column vector of length M is an M × 1 array.A matrix of dimensions M rows and M columns is an M ×N array.

Variable name conventions

MATLAB is case sensitiveStarts with A-Z, a-z

Up to 31 letters, digits and underscores

Default MATLAB variables

ans result of last unassigned calculationeps smallest number that can be added to 1.0

and still be differentInf infinity e.g. 1/0 = Inf

NaN Not a Number e.g. 0/0 = NaN

pi Value of Pi (3.1415...)realmax Largest real number MATLAB can representrealmin Smallest real number MATLAB can represent

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422 CHAPTER 17. SYNTAX REFERENCE FOR MATLAB

User defined variables

clear clear all or selected variables (or functions)from the current workspace

length length of a vector or maximum dimension of an arraysize display dimensions of a particular arraywho display current workspace variable nameswhos display current workspace variable names,

types and associated sizes

Command/Figure Window control

clc clear the command windowclf clear the figure windowfigure start a new figure windowfigure(N) Make figure with index N active. If N is an

integer and figure(N) does not exist, create itclose close current figure windowclose(N) close figure with index N

Useful Workspace functions

help obtain help generally or for a specific functionlookfor obtain one-line help if it existsmore toggles pagination, useful for longs “helps”load read variables in from a disk filesave save all or selected variables to a disk file

MATLAB path and environment

cd change to a specific directorydir list files in the current directorypath display or modify the function search path

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MATLAB punctuation

. decimal point usage: 325/100, 3.25 and .325e1 are all the same

... three or more decimal points at the end of a line causethe following line to be a continuation

, comma is used to separate matrix subscripts and argumentsto functions, also used to separate statements inmulti-statement lines

; used inside brackets to indicate the ends of the rows of amatrix, also used after an expression or statementto suppress printing

% begins comments’ Quote. ’ANY TEXT’ is a vector whose components are the

ASCII codes for the characters. A quote within the text isindicated by two quotes. For example: ’Don”t forget.’

Assignment

Individual elements with a row can be delimited by a comma ora space.

Explicit assignment using ;’s to end rows

a = [1,2,3;4,5,6;7,8,9]

Explicit assignment using “newline” to end rows

a = [1,2,3

4,5,6

7,8,9]

Explicit assignment using continuation lines

b = [1 2 3 4 5 6 ...

7 8 9 10]

Useful ways to initialize variables

linspace(a,b,N)


Linearly spaced intervals between a and b (inclusive)comprised of N points

zeros(m,n) An m by n array of zeroeszeros(n) An n by n array of zeroesones(m,n) An m by n array of onesones(n) An n by n array of oneseye(m,n) An m by n array with ones on the diagonaleye(n) An n by n identity matrixones(n) An n by n array of onesrand(m,n) An m by n array of random numbersrand(n) An n by n array of random numbers

Vector/Matrix operators

These are Vector and Matrix operators

+ addition- subtraction* multiplication/ left division\ right division^ exponentiation’ transpose

Point-by-point operators

These operate on matrix elements in point-wise fashion

.* point-wise multiplication

./ point-wise left division

.\ point-wise right division

.^ point-wise exponentiation

Logical operators

< less than<= less than or equal> greater than

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>= greater than or equal== equal~= not equal& logical AND| logical OR~ logical NOT

List generation/variable indexing

i:k:l List generation syntax = 1stValue:Stride:LatValuev(1) 1st element of vector vv(end) Last element of vector vv(1:2:9) 1st, 3rd, 5th, 7th, 9th elements of vector vv(2:3:9) 2nd, 5th, 8th elements of vector va(2,3) 2nd row, 3rd column of matrix a

a(:,3) all elements in column 3a(1,:) all elements in row 1a(1:2:end,:) all odd rows of matrix a

a(1:2,2:4) sub-matrix of rows 1 and 2, columns 2 through 4a(1,end) last element in 1’st row

Script M-files

Sequences of MATLAB commands can be stored in text files with the extension .m. Thecommands can be executed by typing the name of the files (without the extension) or throughthe file management tools provided by the Command Window menu.

Useful M-file functions

Function M-files

Function definition

Define a separate file called functionName.m with the following form:

function [out1,...,outN] = functionName(in1,...inM)

%FUNCTIONNAME A brief one line description (optional)


% .

% .

% More description (optional)

% .

% .

% .

% First executable statement follows this line

.

.

.

% Valid executable MATLAB statements and comments

.

.

.

% Last line in functionName

Function call

The function call is made with the following statement:

[out1,out2,...,outN] = functionName(in1,in2,...inM)

Useful M-file functions

disp display a stringfprintf write data to screen of fileecho toggle command echoerror display message and abortinput prompt for inputkeyboard transfer control to keyboardpause wait for time or user responsereturn return to callerwarning display warning messages

Performance monitoring

tic,toc stopwatch timer functionsflops counts floating point operations

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Formatting

format short Scaled fixed point format with 5 digitsformat long Scaled fixed point format with 15 digitsformat compact Suppress extra line-feedsformat loose Puts extra line-feeds in the output

Program Flow control

for loops

for k = vectorOrColumnList

% MATLAB statements

end

while loops

while logicalExpression

% MATLAB statements

end

if/elseif/else construct

if logicalExpression1 % Mandatory

% MATLAB statements

elseif logicalExpression2 % Optional

% MATLAB statements

elseif logicalExpression3 % Optional

.

.

.

elseif logicalExpressionN % Optional

% MATLAB statements

else % Optional

% MATLAB statements

end % Mandatory


Plotting

contour Contour plot on a planecontour3 3D contour plot with displayed depthmesh 3D mesh surfacemeshc Combination mesh/contour plotmeshz 3D mesh with curtainpcolor Pseudocolor (checkerboard) plotplot Basic 2D plotsplot3 Plot lines and points in 3D spacesurf 3D colored surfacesurfc Combination surf/contour plotsurfl 3D shaded surface with lighting

Plotting annotation

clabel Contour plot elevation labelscolorbar Display color bar (color scale)legend Graph legendtitle Graph titlexlabel X-axis labelylabel Y-axis label

Additional plotting functions

box Toggle the box displaycolormap Color look-up tablegrid Toggle the grid statehold Control multiple plots on a single figureshading Color shading mode e.g. flat, interp

subplot Control multiple plots in one windowzoom Enable mouse-based zooming

Index

algorithmal-Kashi’s power algorithm, 38, 39Breakfast, 28definition, 1, 23–28design, 29, 38, 39the three pillars of, 26

raising a number to an integral power,38, 39

roots of the quadratic equation, 29

computerarchitecture, 23, 40

flowchart, 24simple branching example, 25simple looping example, 26simple sequence example, 24

Gauss, Karl Friedrich 1777–1855 AD, 35,44

instructionbranch, 25conditional, 25decisions, 25definition, 23flow, 24jump, 25, 26looping, 25, 26sequencing, 23

logicflow, 24

pseudocode, 24simple branching example, 25simple looping example, 26

simple sequence example, 24

sequencing, see instruction, sequencingstatement, see instruction

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Eng Textbook

Documents

ctional binary

dimensional

bit binary

real decimal

bit patterns

bit pattern

bit pattern

signed integer